Problem Consider the following two elements of \(S_{5}\)\begin{align*} g_{1} & =\left [ 54123\right ] \\ g_{2} & =\left [ 21534\right ] \end{align*}
Find a third element \(g\) of this group such that \(g^{-1}g_{1}g=g_{2}\)
Solution
When \(g^{-1}xg=y\), we say that \(y\) is conjugate to \(x\) using \(g\). \begin{align} gg^{-1}g_{1}g & =gg_{2}\nonumber \\ g_{1}g & =gg_{2} \tag{1} \end{align}
But the class of conjugate pairs is symmetric. This means that\begin{align} g^{-1}g_{2}g & =g_{1}\nonumber \\ gg^{-1}g_{2}g & =gg_{1}\nonumber \\ g_{2}g & =gg_{1} \tag{2} \end{align}
We have two equations (1,2). Let us now apply \(g_{1},g_{2}\) on them. Let \(g=\left [ abcde\right ] \) and the goal is to determine the unknowns \(a,b,c,d,e\). Equation (1) becomes\begin{align} \left [ 54123\right ] \left [ abcde\right ] & =\left [ abcde\right ] \left [ 21534\right ] \nonumber \\ \left [ edabc\right ] & =\left [ abcde\right ] \left [ 21534\right ] \tag{1A} \end{align}
Similarly for (2)\begin{align} \left [ 21534\right ] \left [ abcde\right ] & =\left [ abcde\right ] \left [ 54123\right ] \nonumber \\ \left [ baecd\right ] & =\left [ abcde\right ] \left [ 54123\right ] \tag{2A} \end{align}
OK, this is some progress. But how are we going to find \(a,b,c,d,e\,?\). Let use try \(a=1\) and see what we get. If \(a=1\) then (1A) implies \(e=2\) and (2A) implies \(b=5\). Now, if \(b=5\) then (1A) gives \(d=4\) and (2A) gives \(a=3\). Which is conflict with our assumption that \(a=1\) we started with.
Let us next assume that \(a=2\) and see if we get a conflict or not. If \(a=2\) then (1A) gives \(e=1\) and (2A) gives \(b=4\). Now, if \(b=4\) then (1A) gives \(d=3\) and (2A) gives \(a=2\). Good no conflict so far. Now taking \(d=3\) then (1A) gives \(b=5\), which is a conflict of what we found so far. So our starting guess of \(a=2\) is not correct.
Let us next assume that \(a=3\) and see if we get a conflict or not. If \(a=3\) then (1A) gives \(e=5\) and (2A) gives \(b=1\). Now using \(b=1\) then (1A) gives \(d=2\) and (2A) gives \(a=5\), which is conflict with our assumption that \(a=3.\)
Let us next assume that \(a=4\) and see if we get a conflict or not. If \(a=4\) then (1A) gives \(e=3\) and (2A) gives \(b=2\). Now using \(b=2\) then (1A) gives \(d=1\) and (2A) gives \(a=4\). Good. No conflict so far. So far we found \(a,b,e,d=4,2,3,1\). It must mean this case that \(c=5\) since it it only entry left. Let us check if this works or not.
From above we have a candidate element to check which is \[ g=\left [ 42513\right ] \] Trying it on (1,2). From (1)\begin{align*} g_{1}g & =gg_{2}\\ \left [ 54123\right ] \left [ 42513\right ] & =\left [ 42513\right ] \left [ 21534\right ] \\ \left [ 31425\right ] & =\left [ 31425\right ] \end{align*}
OK. Let us check (2)\begin{align*} g_{2}g & =gg_{1}\\ \left [ 21534\right ] \left [ 42513\right ] & =\left [ 42513\right ] \left [ 54123\right ] \\ \left [ 24351\right ] & =\left [ 24351\right ] \end{align*}
Verified. Hence one element is \(g=\left [ 42513\right ] \) .
This means that
\[ \left [ 42513\right ] ^{-1}\left [ 54123\right ] \left [ 42513\right ] =\left [ 21534\right ] \]
Do the following matrices for a group?\[\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} ,\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \] Here \(z=e^{i\frac{2\pi }{3}}\). If not, add the minimum number of \(2\times 2\) matrices to form a group. Then make a list of all possible subgroups.
Solution
The group \(G\) with elements \(g_{i}\) must have the following properties (using matrix multiplication as the binary operation \(\circ \))
Checking the first property. Let \(g_{1}=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} ,g_{2}=\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} ,g_{3}=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \), then since \(g_{1}\) is the identity element, all products with it will also be in \(G\). Looking at products with \(g_{2}\)\begin{align*} g_{2}g_{3} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \\ & =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} \end{align*}
But \(\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} \) is not in \(G\). Hence it is not a group since not closed under the matrix multiplication.
Adding this as new element and calling it \(g_{4}\)\[ g_{4}=\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} \] But now we see that \[ g_{2}g_{4}=\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} \] Is not in \(G\). Calling the \(g_{5}\). \[ g_{5}=\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} \] Check again if closed\begin{align*} g_{2}g_{5} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} \\ & =\begin{pmatrix} 0 & z^{3}\\ z^{3} & 0 \end{pmatrix} =\begin{pmatrix} 0 & e^{i2\pi }\\ e^{i2\pi } & 0 \end{pmatrix} =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =g_{3} \end{align*}
Which is in \(G\). Now checking all products with \(g_{3}\) to see if they are in \(G\).\[ g_{3}g_{2}=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =g_{5}\] Which is in \(G\). And \[ g_{3}g_{4}=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} \] But this is not in \(G\). Adding the above as new element \(g_{6}\)\[ g_{6}=\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} \] Checking again from the start that the group we have now is closed, which now contains \(g_{1},g_{2},g_{3},g_{4},g_{5},g_{6}\).
Checking all products with \(g_{2}\)\begin{align*} g_{2}g_{2} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z^{4}\end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & e^{i\frac{2\pi }{3}\left ( 4\right ) }\end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & e^{i\frac{\pi }{3}}\end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =g_{6}\\ g_{2}g_{3} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =g_{4}\\ g_{2}g_{4} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z^{4} & 0 \end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =g_{5}\\ g_{2}g_{5} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =\begin{pmatrix} 0 & z^{3}\\ z^{3} & 0 \end{pmatrix} =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =g_{3}\\ g_{2}g_{6} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =\begin{pmatrix} e^{i2\pi } & 0\\ 0 & e^{i2\pi }\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =g_{1} \end{align*}
Checking all products with \(g_{3}\)\begin{align*} g_{3}g_{2} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =g_{5}\\ g_{3}g_{3} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =g_{1}\\ g_{3}g_{4} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =g_{6}\\ g_{3}g_{5} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =g_{2}\\ g_{3}g_{6} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =g_{4} \end{align*}
Checking all products with \(g_{4}\)\begin{align*} g_{4}g_{2} & =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix}\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} 0 & z^{3}\\ z^{3} & 0 \end{pmatrix} =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =g_{3}\\ g_{4}g_{3} & =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =g_{2}\\ g_{4}g_{4} & =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =g_{1}\\ g_{4}g_{5} & =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z^{4}\end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & e^{i\frac{2\pi }{3}\left ( 4\right ) }\end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & e^{i\frac{\pi }{3}}\end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =g_{6}\\ g_{4}g_{6} & =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z^{4} & 0 \end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =g_{5} \end{align*}
Checking all products with \(g_{5}\)\begin{align*} g_{5}g_{2} & =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix}\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} 0 & z^{4}\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =g_{4}\\ g_{5}g_{3} & =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =g_{6}\\ g_{5}g_{4} & =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} z^{4} & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =g_{2}\\ g_{5}g_{5} & =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =g_{1}\\ g_{5}g_{6} & =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} 0 & z^{3}\\ z^{3} & 0 \end{pmatrix} =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =g_{3} \end{align*}
Checking all products with \(g_{6}\)\begin{align*} g_{6}g_{2} & =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix}\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =g_{1}\\ g_{6}g_{3} & =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =g_{5}\\ g_{6}g_{4} & =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} 0 & z^{3}\\ z^{3} & 0 \end{pmatrix} =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =g_{3}\\ g_{6}g_{5} & =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =\begin{pmatrix} 0 & z^{4}\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =g_{4}\\ g_{6}g_{6} & =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} z^{4} & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =g_{2} \end{align*}
Therefore the group\[ G=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} ,\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} ,\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} ,\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} \] Is closed under matrix multiplication. To check the associative property, which says that \(g_{i}\circ \left ( g_{j}\circ g_{k}\right ) =\left ( g_{i}\circ g_{j}\right ) \circ g_{k}\) for all \(i,j,k\,\) in \(G\). But from the property of matrix multiplication, we know this property is already satisfied since the matrices are all of same order which is \(2\times 2\). Checking that There is element \(I\) called the identity element such that \(I\circ g_{i}=g_{i}\circ I=g_{i}\), then we see that \(g_{1}=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \) is clearly \(I\) in this case. Checking the last property: Each group element \(g_{i}\) has inverse \(g_{i}^{-1}\) such that \(g_{i}\circ g_{i}^{-1}=g_{i}^{-1}\circ g_{i}=I\). In this case \(g_{i}^{-1}\) is the inverse.
For \(g_{1}\) then \(g_{1}^{-1}\) is itself.
Checking \(g_{2}\)\begin{align*} g_{2}^{-1} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} ^{-1}=\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =g_{6}\\ g_{2}^{-1}\circ g_{2} & =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix}\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =I \end{align*}
Checking \(g_{3}\)\begin{align*} g_{3}^{-1} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} ^{-1}=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =g_{3}\\ g_{3}^{-1}\circ g_{3} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =I \end{align*}
Checking for \(g_{4}\)\begin{align*} g_{4}^{-1} & =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} ^{-1}=\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =g_{4}\\ g_{4}^{-1}\circ g_{4} & =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =I \end{align*}
Checking \(g_{5}\)\begin{align*} g_{5}^{-1} & =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} ^{-1}=\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =g_{5}\\ g_{5}^{-1}\circ g_{5} & =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =I \end{align*}
Checking \(g_{6}\)\begin{align*} g_{6}^{-1} & =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} ^{-1}=\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =g_{2}\\ g_{6}^{-1}\circ g_{6} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =I \end{align*}
OK. All elements checked. Hence \(G\) is indeed a group.\[ G=\overset{g_{1}}{\overbrace{\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} }},\overset{g_{2}}{\overbrace{\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} }},\overset{g_{3}}{\overbrace{\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} }},\overset{g_{4}}{\overbrace{\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} }},\overset{g_{5}}{\overbrace{\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} }},\overset{g_{6}}{\overbrace{\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} }}\] Setting up the Group table. In this table \(g_{1}=I\) the identity element.
| \(\circ \) | \(I\) | \(g_{2}\) | \(g_{3}\) | \(g_{4}\) | \(g_{5}\) | \(g_{6}\) |
| \(I\) | \(I\) | \(g_{2}\) | \(g_{3}\) | \(g_{4}\) | \(g_{5}\) | \(g_{6}\) |
| \(g_{2}\) | \(g_{2}\) | \(g_{6}\) | \(g_{4}\) | \(g_{5}\) | \(g_{3}\) | \(I\) |
| \(g_{3}\) | \(g_{3}\) | \(g_{5}\) | \(I\) | \(g_{6}\) | \(g_{2}\) | \(g_{4}\) |
| \(g_{4}\) | \(g_{4}\) | \(g_{3}\) | \(g_{2}\) | \(I\) | \(g_{6}\) | \(g_{5}\) |
| \(g_{5}\) | \(g_{5}\) | \(g_{4}\) | \(g_{6}\) | \(g_{2}\) | \(I\) | \(g_{3}\) |
| \(g_{6}\) | \(g_{6}\) | \(I\) | \(g_{5}\) | \(g_{3}\) | \(g_{4}\) | \(g_{2}\) |
Now we need to find all subgroups. By Lagrange theorem, we know for finite group such as \(G\) above, all subgroups are of order that divides the order of \(G\). This means the order of the subgroups (if they exist) must be \(2\) or \(3\). (not counting order \(1\) which is just \(I\) and order \(6\) which is the group \(G\) itself).
Let us consider possible subgroups of order \(2\) first. Since subgroup must include the identity element \(g_{1}=I\), then all possible subgroups of order \(2\) are the following
\[ \left [ I,g_{2}\right ] ,\left [ I,g_{3}\right ] ,\left [ I,g_{4}\right ] ,\left [ I,g_{5}\right ] ,\left [ I,g_{6}\right ] \] Clearly each one of these is closed under \(\circ \). Since \(I\circ g_{i}=g_{i}\circ I=g_{i}\in G_{sub}\). But when checking for the property that each group element \(g_{i}\) has inverse \(g_{i}^{-1}\) such that \(g_{i}\circ g_{i}^{-1}=g_{i}^{-1}\circ g_{i}=I\), then this fails unless each element is the same as its inverse. From earlier we found that
\begin{align*} g_{3}^{-1} & =g_{3}\\ g_{4}^{-1} & =g_{4}\\ g_{5}^{-1} & =g_{5} \end{align*}
Only. This implies that out of the above 6 candidate subgroups of order 2 only the following are subgroups\[ \left [ I,g_{3}\right ] ,\left [ I,g_{4}\right ] ,\left [ I,g_{5}\right ] \] We found 3 subgroups so far. Now we need to consider all possible subgroups of order 3. Candidates are\[ \left [ I,g_{2},g_{3}\right ] ,\left [ I,g_{2},g_{4}\right ] ,\left [ I,g_{2},g_{5}\right ] ,\left [ I,g_{2},g_{6}\right ] ,\left [ I,g_{3},g_{4}\right ] ,\left [ I,g_{3},g_{5}\right ] ,\left [ I,g_{3},g_{6}\right ] ,\left [ I,g_{4},g_{5}\right ] ,\left [ I,g_{4},g_{6}\right ] ,\left [ I,g_{5},g_{6}\right ] \] There are \(10\) candidates subgroups of order \(3\) above that we need to check. Easiest check is if the subgroup is closed. We know they satisfy the associative property.
Checking \(\left [ I,g_{2},g_{3}\right ] \)\[ g_{2}\circ g_{3}=\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =g_{4}\] Not closed.
Checking \(\left [ I,g_{2},g_{4}\right ] \)\[ g_{2}\circ g_{4}=\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =g_{5}\] Not closed.
Checking \(\left [ I,g_{2},g_{5}\right ] \)\[ g_{2}\circ g_{5}=\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =g_{3}\] Not closed.
Checking \(\left [ I,g_{2},g_{6}\right ] \)\begin{align*} g_{2}\circ g_{6} & =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =I\\ g_{6}\circ g_{2} & =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix}\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =\begin{pmatrix} z^{3} & 0\\ 0 & z^{3}\end{pmatrix} =I \end{align*}
Closed. Associativity is met since these are matrices of same order. Let check inverse property: Each subgroup element \(g_{i}\) has inverse \(g_{i}^{-1}\) such that \(g_{i}\circ g_{i}^{-1}=g_{i}^{-1}\circ g_{i}=I\). In this case \(g_{i}^{-1}\) is the inverse matrix.
For \(g_{2}\)\[ g_{2}^{-1}=\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} ^{-1}=\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =g_{6}\] And \(g_{2}^{-1}\circ g_{2}=I\). OK. And\[ g_{6}^{-1}=\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} ^{-1}=\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =g_{2}\] And \(g_{6}^{-1}\circ g_{6}=I\). OK. Therefore \(\left [ I,g_{2},g_{6}\right ] \) is indeed a subgroup.
Checking \(\left [ I,g_{3},g_{4}\right ] \)\[ g_{3}\circ g_{4}=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =g_{6}\] But \(g_{6}\) is not in this subgroup. Hence not closed.
Checking \(\left [ I,g_{3},g_{5}\right ] \)\[ g_{3}\circ g_{5}=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} =g_{2}\] But \(g_{2}\) is not in this subgroup. Hence not closed.
Checking \(\left [ I,g_{3},g_{6}\right ] \)\[ g_{3}\circ g_{6}=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} =g_{4}\] But \(g_{4}\) is not in this subgroup. Hence not closed.
Checking \(\left [ I,g_{4},g_{5}\right ] \)\[ g_{4}\circ g_{5}=\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix}\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =g_{6}\] But \(g_{5}\) is not in this subgroup. Hence not closed.
Checking \(\left [ I,g_{4},g_{6}\right ] \)\[ g_{4}\circ g_{6}=\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z^{4} & 0 \end{pmatrix} =\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} =g_{5}\] But \(g_{5}\) is not in this subgroup. Hence not closed.
Checking \(\left [ I,g_{5},g_{6}\right ] \)\[ g_{5}\circ g_{6}=\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix}\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} =\begin{pmatrix} 0 & z^{3}\\ z^{3} & 0 \end{pmatrix} =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =g_{3}\] But \(g_{3}\) is not in this subgroup. Hence not closed.
All subgroups of order \(3\) are checked. Therefore the following are the subgroups found. There are \(4\) in total \[ \left [ I,g_{3}\right ] ,\left [ I,g_{4}\right ] ,\left [ I,g_{5}\right ] ,\left [ I,g_{2},g_{6}\right ] \] Or\begin{align*} & \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \\ & \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} 0 & z\\ z^{2} & 0 \end{pmatrix} \\ & \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} 0 & z^{2}\\ z & 0 \end{pmatrix} \\ & \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} z & 0\\ 0 & z^{2}\end{pmatrix} ,\begin{pmatrix} z^{2} & 0\\ 0 & z \end{pmatrix} \end{align*}
The Lorentz transformation with velocity \(v\) along the \(x\) axis is described by\[\begin{pmatrix} x^{\prime }\\ t^{\prime }\end{pmatrix} =M\left ( v\right ) \begin{pmatrix} x\\ t \end{pmatrix} \] Where \(M\left ( v\right ) =\frac{1}{\sqrt{1-v^{2}}}\begin{pmatrix} 1 & v\\ v & 1 \end{pmatrix} \). Show that the product of two such transformations is again a Lorentz transformation. i.e. \(M\left ( v_{2}\right ) M\left ( v_{1}\right ) =M\left ( v_{12}\right ) \) and find \(v_{12}\). Using this result, show that these transformations form a group.
solution
The following diagram is used to help in understanding what we are trying to show.
Given\[\begin{pmatrix} x^{\prime }\\ t^{\prime }\end{pmatrix} =M\left ( v_{1}\right ) \begin{pmatrix} x\\ t \end{pmatrix} \] And\[\begin{pmatrix} x^{\prime \prime }\\ t^{\prime \prime }\end{pmatrix} =M\left ( v_{2}\right ) \begin{pmatrix} x^{\prime }\\ t^{\prime }\end{pmatrix} \] We need to show that, with the help of the diagram above, that\[\begin{pmatrix} x^{\prime \prime }\\ t^{\prime \prime }\end{pmatrix} =M\left ( v_{2}\right ) \begin{pmatrix} x^{\prime }\\ t^{\prime }\end{pmatrix} =M\left ( v_{2}\right ) M\left ( v_{1}\right ) \begin{pmatrix} x\\ t \end{pmatrix} =M\left ( v_{12}\right ) \begin{pmatrix} x\\ t \end{pmatrix} \] So we need to find \(M\left ( v_{12}\right ) \) and see if it is a Lorentz transformation also. In other words, to see if \(M\left ( v_{12}\right ) \) has the form of\(\frac{1}{\sqrt{1-v_{12}^{2}}}\begin{pmatrix} 1 & v_{12}\\ v_{12} & 1 \end{pmatrix} \) and need to find what \(v_{12}\) is. Starting by finding \(M\left ( v_{2}\right ) \). Given that\begin{align*} \begin{pmatrix} x^{\prime }\\ t^{\prime }\end{pmatrix} & =M\left ( v_{1}\right ) \begin{pmatrix} x\\ t \end{pmatrix} \\ & =\frac{1}{\sqrt{1-v_{1}^{2}}}\begin{pmatrix} 1 & v_{1}\\ v_{1} & 1 \end{pmatrix}\begin{pmatrix} x\\ t \end{pmatrix} \\ & =\frac{1}{\sqrt{1-v_{1}^{2}}}\begin{pmatrix} x+v_{1}t\\ v_{1}x+t \end{pmatrix} \end{align*}
The above gives\begin{align*} x^{\prime } & =\frac{1}{\sqrt{1-v_{1}^{2}}}\left ( x+v_{1}t\right ) \\ t^{\prime } & =\frac{1}{\sqrt{1-v_{1}^{2}}}\left ( v_{1}x+t\right ) \end{align*}
Applying the transformation again on the above result gives\begin{align*} \begin{pmatrix} x^{\prime \prime }\\ t^{\prime \prime }\end{pmatrix} & =M\left ( v_{2}\right ) \begin{pmatrix} x^{\prime }\\ t^{\prime }\end{pmatrix} \\ & =\frac{1}{\sqrt{1-v_{2}^{2}}}\begin{pmatrix} 1 & v_{2}\\ v_{2} & 1 \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{1-v_{1}^{2}}}\left ( x+v_{1}t\right ) \\ \frac{1}{\sqrt{1-v_{1}^{2}}}\left ( v_{1}x+t\right ) \end{pmatrix} \\ & =\frac{1}{\sqrt{1-v_{2}^{2}}}\begin{pmatrix} \frac{1}{\sqrt{1-v_{1}^{2}}}\left ( x+v_{1}t\right ) +\frac{v_{2}}{\sqrt{1-v_{1}^{2}}}\left ( v_{1}x+t\right ) \\ \frac{v_{2}}{\sqrt{1-v_{1}^{2}}}\left ( x+v_{1}t\right ) +\frac{1}{\sqrt{1-v_{1}^{2}}}\left ( v_{1}x+t\right ) \end{pmatrix} \\ & =\frac{1}{\sqrt{1-v_{2}^{2}}}\frac{1}{\sqrt{1-v_{1}^{2}}}\begin{pmatrix} \left ( x+v_{1}t\right ) +v_{2}\left ( v_{1}x+t\right ) \\ v_{2}\left ( x+v_{1}t\right ) +\left ( v_{1}x+t\right ) \end{pmatrix} \\ & =\frac{1}{\sqrt{\left ( 1-v_{2}^{2}\right ) \left ( 1-v_{1}^{2}\right ) }}\begin{pmatrix} x+v_{1}t+v_{2}v_{1}x+v_{2}t\\ v_{2}x+v_{2}v_{1}t+v_{1}x+t \end{pmatrix} \\ & =\frac{1}{\sqrt{1-v_{1}^{2}-v_{2}^{2}+v_{2}^{2}v_{1}^{2}}}\begin{pmatrix} x\left ( 1+v_{2}v_{1}\right ) +t\left ( v_{1}+v_{2}\right ) \\ x\left ( v_{2}+v_{1}\right ) +t\left ( 1+v_{2}v_{1}\right ) \end{pmatrix} \\ & =\frac{\left ( 1+v_{2}v_{1}\right ) }{\sqrt{1-v_{1}^{2}-v_{2}^{2}+v_{2}^{2}v_{1}^{2}}}\begin{pmatrix} x+t\frac{\left ( v_{1}+v_{2}\right ) }{\left ( 1+v_{2}v_{1}\right ) }\\ x\frac{\left ( v_{2}+v_{1}\right ) }{\left ( 1+v_{2}v_{1}\right ) }+t \end{pmatrix} \\ & =\frac{1}{\sqrt{\frac{1-v_{1}^{2}-v_{2}^{2}+v_{2}^{2}v_{1}^{2}}{\left ( 1+v_{2}v_{1}\right ) ^{2}}}}\begin{pmatrix} 1 & \frac{\left ( v_{1}+v_{2}\right ) }{\left ( 1+v_{2}v_{1}\right ) }\\ \frac{\left ( 1+v_{2}v_{1}\right ) }{\left ( 1+v_{2}v_{1}\right ) } & 1 \end{pmatrix}\begin{pmatrix} x\\ t \end{pmatrix} \end{align*}
But \[ \frac{1-v_{1}^{2}-v_{2}^{2}+v_{2}^{2}v_{1}^{2}}{\left ( 1+v_{2}v_{1}\right ) ^{2}}=\frac{\left ( 1+v_{1}v_{2}\right ) ^{2}-\left ( v_{1}+v_{2}\right ) ^{2}}{\left ( 1+v_{2}v_{1}\right ) ^{2}}\] Therefore\begin{equation} \begin{pmatrix} x^{\prime \prime }\\ t^{\prime \prime }\end{pmatrix} =\frac{1}{\sqrt{1-\frac{\left ( v_{1}+v_{2}\right ) ^{2}}{\left ( 1+v_{2}v_{1}\right ) ^{2}}}}\begin{pmatrix} 1 & \frac{\left ( v_{1}+v_{2}\right ) }{\left ( 1+v_{2}v_{1}\right ) }\\ \frac{\left ( 1+v_{2}v_{1}\right ) }{\left ( 1+v_{2}v_{1}\right ) } & 1 \end{pmatrix}\begin{pmatrix} x\\ t \end{pmatrix} \tag{1} \end{equation} Now it is in the form of Lorentz transformation. \(\begin{pmatrix} x^{\prime \prime }\\ t^{\prime \prime }\end{pmatrix} =M\left ( v_{12}\right ) \begin{pmatrix} x\\ t \end{pmatrix} \). Comparing this (1) shows that\[ M\left ( v_{12}\right ) =\frac{1}{\sqrt{1-\frac{\left ( v_{1}+v_{2}\right ) ^{2}}{\left ( 1+v_{2}v_{1}\right ) ^{2}}}}\begin{pmatrix} 1 & \frac{\left ( v_{1}+v_{2}\right ) }{\left ( 1+v_{2}v_{1}\right ) }\\ \frac{\left ( 1+v_{2}v_{1}\right ) }{\left ( 1+v_{2}v_{1}\right ) } & 1 \end{pmatrix} \] But \(M\left ( v\right ) =\frac{1}{\sqrt{1-v^{2}}}\begin{pmatrix} 1 & v\\ v & 1 \end{pmatrix} \). By comparing to the above shows that \[ v_{12}=\frac{v_{1}+v_{2}}{1+v_{2}v_{1}}\] Therefore what we did above is apply Lorentz transformation again \(M\left ( v_{2}\right ) \) on result we obtained from \(M\left ( v_{1}\right ) \) and we obtained a result which also a valid Lorentz transformation. This means the group is closed under this transformation. We need to show associativity. Which means\begin{align} M\left ( v_{3}\right ) \left [ M\left ( v_{2}\right ) M\left ( v_{1}\right ) \right ] & =\left [ M\left ( v_{3}\right ) M\left ( v_{2}\right ) \right ] M\left ( v_{1}\right ) \nonumber \\ M\left ( v_{3}\right ) M\left ( v_{12}\right ) & =M\left ( v_{23}\right ) M\left ( v_{1}\right ) \tag{3} \end{align}
But we found from the above that \(M\left ( v_{2}\right ) M\left ( v_{1}\right ) =M\left ( v_{12}\right ) \) results in \(v_{12}=\frac{v_{1}+v_{2}}{1+v_{2}v_{1}}\). Therefore we can conclude that left side of (2) which is \(M\left ( v_{3}\right ) M\left ( v_{12}\right ) \) will also result in\[ v_{321}=\frac{v_{12}+v_{3}}{1+v_{3}v_{12}}\] But \(v_{12}=\frac{v_{1}+v_{2}}{1+v_{2}v_{1}}\), therefore the above simplifies to\begin{align} v_{321} & =\frac{\frac{v_{1}+v_{2}}{1+v_{2}v_{1}}+v_{3}}{1+v_{3}\frac{v_{1}+v_{2}}{1+v_{2}v_{1}}}\nonumber \\ & =\frac{v_{1}+v_{2}+v_{3}\left ( 1+v_{2}v_{1}\right ) }{1+v_{2}v_{1}+v_{3}v_{1}+v_{2}}\nonumber \\ & =\frac{v_{1}+v_{2}+v_{3}+v_{3}v_{2}v_{1}}{1+v_{2}v_{1}+v_{3}v_{1}+v_{3}v_{2}} \tag{3A} \end{align}
And the right side of (3) which is \(M\left ( v_{23}\right ) M\left ( v_{1}\right ) \) also gives\[ v_{123}=\frac{v_{1}+v_{23}}{1+v_{1}v_{23}}\] But again, \(v_{23}=\frac{v_{2}+v_{3}}{1+v_{3}v_{2}}\) and the above simplifies to \begin{align} v_{123} & =\frac{v_{1}+\frac{v_{2}+v_{3}}{1+v_{3}v_{2}}}{1+v_{1}\frac{v_{2}+v_{3}}{1+v_{3}v_{2}}}\nonumber \\ & =\frac{v_{1}\left ( 1+v_{3}v_{2}\right ) +v_{2}+v_{3}}{1+v_{3}v_{2}+v_{1}v_{2}+v_{1}v_{3}}\nonumber \\ & =\frac{v_{1}+v_{3}v_{2}v_{1}+v_{2}+v_{3}}{1+v_{3}v_{2}+v_{1}v_{2}+v_{1}v_{3}} \tag{3B} \end{align}
By comparing (3A) and (3B) we see they are the same. Hence associativity is satisfied. Next we need to check the inverse property. What this means that for each \(M\left ( v_{i}\right ) \) there exist \(M^{-1}\left ( v_{i}\right ) \) such that \(M\left ( v_{i}\right ) M^{-1}\left ( v_{i}\right ) =I\). where the identity in this case is \(M\left ( 0\right ) =I\) since \begin{align*} M\left ( 0\right ) & =\frac{1}{\sqrt{1-0}}\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \\ & =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \end{align*}
Since \(M\left ( v\right ) =\frac{1}{\sqrt{1-v^{2}}}\begin{pmatrix} 1 & v\\ v & 1 \end{pmatrix} \) then \(M\left ( -v\right ) =\frac{1}{\sqrt{1-v^{2}}}\begin{pmatrix} 1 & -v\\ -v & 1 \end{pmatrix} \) and\begin{align*} M\left ( v\right ) M\left ( -v\right ) & =\frac{1}{\sqrt{1-v^{2}}}\begin{pmatrix} 1 & v\\ v & 1 \end{pmatrix} \frac{1}{\sqrt{1-v^{2}}}\begin{pmatrix} 1 & -v\\ -v & 1 \end{pmatrix} \\ & =\frac{1}{1-v^{2}}\begin{pmatrix} 1 & v\\ v & 1 \end{pmatrix}\begin{pmatrix} 1 & -v\\ -v & 1 \end{pmatrix} \\ & =\frac{1}{1-v^{2}}\begin{pmatrix} 1-v^{2} & 0\\ 0 & 1-v^{2}\end{pmatrix} \\ & =\frac{1-v^{2}}{1-v^{2}}\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \\ & =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \\ & =M\left ( 0\right ) \end{align*}
Which is the identity. Hence we showed that for each \(M\left ( v_{i}\right ) \) there exists an inverse \(M\left ( -v_{i}\right ) \). All properties of group have been satisfied. Hence the given Lorentz transformation forms a group.
Using \(\left [ X_{i},X_{j}\right ] =c_{ij}^{k}X_{k}\) where \(c_{ij}^{k}\) are the structure constants and a summation over \(k\) is implied.
Conditions (1,3) are the only conditions on the structure constants. Any set of real numbers \(c_{ij}^{k}\) obeying these two conditions defines a Lie algebra.
solution
The commutator of 2 generators (\(X_{i},X_{j}\)) is linear combination of the generators. Hence\begin{equation} \left [ X_{i},X_{j}\right ] =X_{i}X_{j}-X_{j}X_{i}=c_{ij}^{k}X_{k}\tag{1} \end{equation} Therefore, we also have\begin{equation} \left [ X_{j},X_{i}\right ] =X_{j}X_{i}-X_{i}X_{j}=c_{ji}^{k}X_{k}\tag{2} \end{equation} Adding (1) and (2) gives\begin{align*} \left ( X_{i}X_{j}-X_{j}X_{i}\right ) +\left ( X_{j}X_{i}-X_{i}X_{j}\right ) & =c_{ij}^{k}X_{k}+c_{ji}^{k}X_{k}\\ 0 & =X_{k}\left ( c_{ij}^{k}+c_{ji}^{k}\right ) \\ 0 & =c_{ij}^{k}+c_{ji}^{k}\\ c_{ij}^{k} & =-c_{ji}^{k} \end{align*}
Applying the commutator relation \[ \left [ X_{i},X_{j}\right ] =X_{i}X_{j}-X_{j}X_{i}\] Let LHS of the Jacobi identity be \(\Delta \). Applying the above to each term in \(\Delta \) gives\begin{equation} \Delta =\left [ \left ( X_{i}X_{j}-X_{j}X_{i}\right ) ,X_{k}\right ] +\left [ \left ( X_{j}X_{k}-X_{k}X_{j}\right ) ,X_{i}\right ] +\left [ \left ( X_{k}X_{i}-X_{i}X_{k}\right ) ,X_{j}\right ] \tag{1} \end{equation} We want to show that \(\Delta =0\). Now, applying commutator relation again each term of the above gives for the first term\begin{align} \left [ \left ( X_{i}X_{j}-X_{j}X_{i}\right ) ,X_{k}\right ] & =\left ( X_{i}X_{j}-X_{j}X_{i}\right ) X_{k}-X_{k}\left ( X_{i}X_{j}-X_{j}X_{i}\right ) \nonumber \\ & =X_{i}X_{j}X_{k}-X_{j}X_{i}X_{k}-X_{k}X_{i}X_{j}+X_{k}X_{j}X_{i}\tag{2} \end{align}
And for the second term in (1)\begin{align} \left [ \left ( X_{j}X_{k}-X_{k}X_{j}\right ) ,X_{i}\right ] & =\left ( X_{j}X_{k}-X_{k}X_{j}\right ) X_{i}-X_{i}\left ( X_{j}X_{k}-X_{k}X_{j}\right ) \nonumber \\ & =X_{j}X_{k}X_{i}-X_{k}X_{j}X_{i}-X_{i}X_{j}X_{k}+X_{i}X_{k}X_{j}\tag{3} \end{align}
And for the third term in (1)\begin{align} \left [ \left ( X_{k}X_{i}-X_{i}X_{k}\right ) ,X_{j}\right ] & =\left ( X_{k}X_{i}-X_{i}X_{k}\right ) X_{j}-X_{j}\left ( X_{k}X_{i}-X_{i}X_{k}\right ) \nonumber \\ & =X_{k}X_{i}X_{j}-X_{i}X_{k}X_{j}-X_{j}X_{k}X_{i}+X_{j}X_{i}X_{k}\tag{4} \end{align}
Substituting (2,3,4) back into (1) gives\begin{align*} \Delta & =\left ( X_{i}X_{j}X_{k}-X_{j}X_{i}X_{k}-X_{k}X_{i}X_{j}+X_{k}X_{j}X_{i}\right ) \\ & +\left ( X_{j}X_{k}X_{i}-X_{k}X_{j}X_{i}-X_{i}X_{j}X_{k}+X_{i}X_{k}X_{j}\right ) \\ & +\left ( X_{k}X_{i}X_{j}-X_{i}X_{k}X_{j}-X_{j}X_{k}X_{i}+X_{j}X_{i}X_{k}\right ) \end{align*}
We see that all terms cancel each other. Hence \(\Delta =0\) which is what we wanted to show.
The Jacobi identity is\[ \left [ \left [ X_{i},X_{j}\right ] ,X_{k}\right ] +\left [ \left [ X_{j},X_{k}\right ] ,X_{i}\right ] +\left [ \left [ X_{k},X_{i}\right ] ,X_{j}\right ] =0 \] Applying \(\left [ X_{i},X_{j}\right ] =c_{ij}^{l}X_{l}\) on each term in the LHS above gives, where the summation index \(l\) is used in each term, which is OK to do since the terms are separated from each others\begin{align*} 0 & =\left [ c_{ij}^{l}X_{l},X_{k}\right ] +\left [ c_{jk}^{l}X_{l},X_{i}\right ] +\left [ c_{ki}^{l}X_{l},X_{j}\right ] \\ & =c_{ij}^{l}\left [ X_{l},X_{k}\right ] +c_{jk}^{l}\left [ X_{l},X_{i}\right ] +c_{ki}^{l}\left [ X_{l},X_{j}\right ] \end{align*}
Now, applying \(\left [ X_{i},X_{j}\right ] =c_{ij}^{m}X_{m}\) again on each term above and now using \(m\) as the summation index gives\begin{align*} 0 & =c_{ij}^{l}c_{lk}^{m}X_{m}+c_{jk}^{l}c_{li}^{m}X_{m}+c_{ki}^{l}c_{lj}^{m}\\ & =\left ( c_{ij}^{l}c_{lk}^{m}+c_{jk}^{l}c_{li}^{m}+c_{ki}^{l}c_{lj}^{m}\right ) X_{m}\\ & =c_{ij}^{l}c_{lk}^{m}+c_{jk}^{l}c_{li}^{m}+c_{ki}^{l}c_{lj}^{m} \end{align*}
Which is what the problem asked to show.