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As $PSL(2,7)$ has a subgroup of index $7$, and $PSL(2,7)$ is simple, hence it can be embedded in $A_7$. How many copies of $PSL(2,7)$ are in $A_7$?

(There should be at-least 15 copies: If $H\leq A_7$, $H\cong PSL(2,7)$, then $|A_7\colon H|=15$. If we count number of distinct conjugates of $H$ in $A_7$ then we get (some) copies of $PSL(2,7)$ in $A_7$, and this is equal to the index of normalizer of $H$ in $A_7$. As $A_7$ is simple, there must be more than one copies of $PSL(2,7)$ in $A_7$. If $|A_7\colon H|\in \{3,5\}$ then we will have injective homomorphism from $A_7$ into $S_3$ or $S_5$, contradiction. Hence $|N_{A_7}(H)\colon H|=15$, so we have at-least 15 copies of $PSL(2,7)$ in $A_7$.)

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To finish the argument, notice that every copy of H in S7 is conjugate, because H has only one conjugacy class of subgroups of order 7 (Sylow's theorem). The normalizer of a subgroup in A7 versus S7 is pretty well-behaved (same or doubled): in this case the normalizer is the same, and so the index doubles and there are 30 conjugates of H in S7, which split into two classes of 15 in A7. –  Jack Schmidt Apr 20 '11 at 14:11
    
The tricky part is to see why the normalizer of $PSL_2(7)$ in $S_7$ is the same as in $A_7$. This boils down to checking that $PGL_2(7)$ has no subgroup of index $7$, or that $S_4$ inside $PSL_2(7)$ does not extend to $PGL_2(7)$... –  Anvita Apr 20 '11 at 15:21

1 Answer 1

A quick look at the ATLAS shows that $A_7$ has two conjugacy classes (of size 15 each) of subgroups isomorphic to $PSL_2(7)$ which fuse into one class in $S_7$. So, there are a total of 30 subgroups.

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