$PSL(2,7)$ as a subgroup of $A_7$ How can we show that the subgroup of $A_7$ generated by the permutations $x= (1234567)$ and $b=(26)(34)$ has order $168$?
 A: Assuming you want to avoid using a computer, you could observe that your two permutations are both automorphisms of the projective plane with 7 points and lines $\{1,2,6\}$, $\{1,3,4\}$, $\{1,5,7\}$, $\{2,3,7\}$, $\{2,4,5\}$, $\{3,5,6\}$, $\{4,6,7\}$. That would give you an upper bound of 168 on the order.
To get a lower bound, you could note for example that $b$ has conjugates $(37)(45)$ and $(25)(67)$ under powers of $a$, and then try and show that these
two permutations together with $b$ generate a subgroup of order at least 24 stabilizing 1.
A: For reference, this is Exercise 15 in Section 2.8 of Hungerford's graduate Algebra textbook. As such, there should be an answer that requires only the tools developed in the book up to that point. In particular,
here is a proof that doesn't rely on knowledge of the projective plane. 
Let $a=(1234567)$ and $b=(26)(34)$ and let 
 $G = \langle a, b \rangle$. We want to show $|G|=168$. Because $|G|$ divides
 $7! / 2$, we must have $$|G| = 2^{m_2} 3^{m_3}5^{m_5}7^{m_7}$$
where $0 \le m_2 \le 3$, $0 \le m_3 \le 2$, $0 \le m_5 \le 1$, and
$0 \le m_7 \le 1$.
Because $|a|=7$, we must have $7 \mid |G|$, and so $m_7=1$. Let $C_1 = \langle a\rangle$ and note that $C_1 \in Syl_7(G)$
Let us find $n_7$, the number of conjugates of $C_1$ in $G$. Observe
\begin{equation*}
b a b^{-1} = 
(1 6 4 3 5 2 7)  \notin C_1,
\end{equation*}
 giving the conjugate subgroup
\begin{equation*}
C_2 = \langle (1 6 4 3 5 2 7) \rangle = \langle (1 2 3 6 7 5 4) \rangle \ne C_1.
\end{equation*}
The group $G$ acts transitively on $Syl_7(G)$ by conjugation,
 and this induces a homomorphism
$\varphi \colon G \to S_{n_7}$. Note $a C_2 a^{-1} \ne C_2$, and so
 $\varphi(a)$
is not trivial. Hence $|\varphi(a)|$ must be $7$. Because this is prime, $\varphi(a)$
must be a product of disjoint 7-cycles, one of which moves $C_2$.
Thus the iteration $C_{i+1} = a C_i a^{-1}$ will give six more Sylow 7-subgroups $C_3, \dotsc, C_8$;
of course, $a C_8 a^{-1} = C_2$ again. I leave these calculations to you. (You need only calculate generators. No need to list out all the elements of each subgroup, because because every elements is a power of the generator.) Becasue $|b|=2$, $\varphi(b)$
is a product of disjoint 2-cylces. We calculate 
$b C_3 b^{-1} = C_8$, $b C_4 b^{-1} = C_5$, and $b C_6 b^{-1} = C_7$;
by definition we also had $b C_1 b^{-1} = C_2$. Because $a$ and $b$ generate $G$, this proves that the orbit of $C_1$ is $\{ C_1, \dotsc, C_8\}$. But the action of
$G$ on $Syl_7 (G)$ is transitive, and so 
$Syl_7(G) = \{C_1, \dotsc, C_8\}$.
Therefore $n_7=8$. But $n_7 \mid |G|$ by the Third Sylow Theorem, and so
 $m_2 = 3$.
Next observe that the Sylow 3-subgroups of $A_7$ are all of the form 
$\langle \sigma_1, \sigma_2 \rangle$ where $\sigma_1, \sigma_2$ are disjoint
3-cycles. Now, 
\begin{equation*}
a b = (1 2 7)(3 5 6) \in \langle (127), (356) \rangle\in Syl_3 (A_7).
\end{equation*}
Becasue $|ab|=3$, we must have $m_3 \ge 1$. Suppose for contradiction
that $m_3 = 2$.
Then $ab$ would have to be contained in some Sylow 3-subgroup of $G$,
but the only group of order 9 in $A_7$ containing $ab$ is 
$  \langle {(127), (356)} \rangle$. Thus $(127) \in G$.
If we conjugate $a$ by $(127)$, the result
$(1273456)$
must lie in $G$.
Because that is not among the elements of
$C_1, \dotsc, C_8$ calculated above, we have a contradiction. Thus 
$m_3 =1$.
Finally, suppose for contradiction that $m_5=1$.
From our results so far we can conclude 
$[A_7 : G]=6$. But $A_7$ is simple,
 so that $A_7$ must be isomorphic to a subgroup of $S_6$ (by considering the action of $A_7$ on the left cosets of $G$).
 But this is impossible becasue $S_6$ has no elements of order 7.
 Thus $m_5 \ne 1$, and so $m_5 = 0$.
Therefore $|G| = 2^3 \cdot 3^1 \cdot 5^0 \cdot 7^1 = 168$.
