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How can we show that the subgroup of $A_7$ generated by the permutations $x= (1234567)$ and $b=(26)(34)$ has order $168$?

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1 Answer 1

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.

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