Group generated by $(1234567)$ and $(26)(34)$ is of order 168.

Question

Let $G$ be a subgroup of $S_7$ generated by $(1234567)$ and $(26)(34)$. Show that $|G| = 168$.

This is a question from Algebra by Hungerford (page.112, exercise 15). And I just have no idea of dealing with this kind of problem about the structure of large number finite groups. Could you please also recommend some reference books?

Answer 1

Here is a short outline.

Set $x=(1 2 3 4 5 6 7)$, $y=(2 6)(3 4)$ and $S = <x>$. Clearly $G = <x, y> \leq Alt(7)$. Hence $S$ is a Sylow $7$-subgroup of $G$. $Alt(7)$ has 6! elements of order $7$ and therefore 5! Sylow $7$-subgroups. Hence $|N_{Alt(7)}(S)| = 21$. Set $T_1 = S^{y}$ and $T_{i+1} = T_{i}^{x}$. The set $\{S, T_{1}, \ldots, T_{7}\}$ is $G$-invariant. Hence it constitutes the set of all Sylow $7$-subgroups of $G$. In particular $|G : N_{G}(S)| = 8$. Since $S \leq N_{G}(S) \leq N_{Alt(7)}(S)$ either $|N_{G}(S)| = 7$ or $|N_{G}(S)| = 21$. As a result there is only two options for $|G|$. Now determine the order of $x \cdot y$.

Answer 2

And now for something really out there...

Consider the seven nonzero points in $F_2^3$, where $F_2$ is the field with three elements. Each element of the 168-element simple group $GL_3(F_2)$ acts on these seven points as a permutation. If we can realize these two elements as invertible matrices, $G$ will be a subgroup of the full linear group, with order dividing $168$.

That simple group is part of the $PSL_n(K)$ family; working over $F_2$ allows us to ignore both the "special" (determinant $1$) and "projective" (quotient by the constant multiples of $I$) parts of that.

So then, first, how do we get something of order $7$? Well, $F_2^3$ can be given a multiplication operator to make it a field. The nonzero elements in that field form a group of order $7$, and any of them other than the identity will have order $7$, acting on the points as a $7$-cycle. Let $x$ be one of these nontrivial elements, and choose a basis $(1,x,x^2)$ for $F_2^3$. There is some polynomial relation writing $x^3$ as a linear combination of the smaller ones - this will be an irreducible polynomial, the minimal polynomial of $x$. We have two choices - either $x^3+x^2+1=0$ or $x^3+x+1=0$. Let's go with the former. Then, with respect to that basis, multiplication by $x$ has the matrix $X=\begin{bmatrix}0&0&1\\1&0&0\\0&1&1\end{bmatrix}$. The seven elements we cycle through are, in order, $1=\begin{bmatrix}1\\0\\0\end{bmatrix}$, $x=\begin{bmatrix}0\\1\\0\end{bmatrix}$, $x^2=\begin{bmatrix}0\\0\\1\end{bmatrix}$, $x^3=\begin{bmatrix}1\\0\\1\end{bmatrix}$, $x^4=\begin{bmatrix}1\\1\\1\end{bmatrix}$, $x^5=\begin{bmatrix}1\\1\\0\end{bmatrix}$, and $x^6=\begin{bmatrix}0\\1\\1\end{bmatrix}$.

OK, now how can we get that $(26)(34)$ permutation? Which element is labeled "1" is an arbitrary choice, so we can start that anywhere in the cycle. For simplicity, make those labels the exponent of $x$ ($1=x^0=x^7$). So then, we need an invertible matrix that fixes $\begin{bmatrix}1\\0\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\\0\end{bmatrix}$, while sending $\begin{bmatrix}0\\0\\1\end{bmatrix}$ to $\begin{bmatrix}0\\1\\1\end{bmatrix}$. That matrix is $A=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}$. Checking its action on the other elements - $Ax^3=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}=x^4$, $Ax^4=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\0\\1\end{bmatrix}=x^3$, $Ax^5=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}1\\1\\0\end{bmatrix}=x^5$, and $Ax^6=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix}=x^2$. Confirmed; these matrices have the proper action, and we have realized this group of permutations as a subgroup of $SL_3(F_2)$.

So then, all that remains is to show that the group generated by these two matrices $A$ and $X$ is the whole linear group $G=SL_3(F_2)$. To make our lives easier, we note that $G$ cannot have any nontrivial subgroups with index $\le 5$; if $H$ is a subgroup with index $k$, $G$ acts on the conjugates of $H$ by conjugation, giving a homomorphism from $G$ to the symmetric group on $k$ elements. The kernel of that is a normal subgroup with index at most $k!$ in $G$, the intersection of all conjugates of $H$. Since we know $G$ is simple, it can't have any nontrivial normal subgroups of order $\le 5!=120$, and there are no nontrivial subgroups of index $\le 5$.
In particular, there are no subgroups of index $2$ or $4$. If we can find an element of order $3$ in the subgroup generated by $A$ and $X$, that subgroup will have order divisible by $2\cdot 3\cdot 7 = \frac{168}{4}$, and will have to be all of $G$.

This is easy. $XA=\begin{bmatrix}0&0&1\\1&0&0\\0&1&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix} =\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$ has order $3$, and the subgroup generated by $A$ and $X$ must be all of $SL_3(F_2)$. We have found not only the group's order but it's full structure, the unique simple group of order $168$.