# Maximal subgroups of of the alternating group of degree 5

$A_5$ has the maximal subgroup $S=<(123),(12)(45)>=\{e,(123),(132),(12)(45),(13)(45),(23)(45)\}$. $A_5$ has $6$ maximal subgroups of order $10$ and has $5$ maximal subgroups of order $12$. I checked with maple software that the intersection of $S$ with any subgroup of order $10$ is nontrivial($\neq 1$). Also, I checked that the intersection of $S$ with any subgroup of order $12$ is nontrivial($\neq 1$).

My question is the following:

Is it possible to prove that the intersection is nontrivial without need for calculation?

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To do the order 10 one (assuming that an order 10 group exists), note that a subgroup of order 10 contains a 5-cycle and an element $\tau$ of order 2 (which is a product of two transpositions). Conjugate $\tau$ by the powers of the 5-cycle to get five different products of two transpositions.

If any two conjugates contain the same transposition, their product is a 3-cycle, so all the transpositions have to be different.

The five elements of order 2 contain 10 transpositions between them. There are only 10 transpositions available, so one of them must be (4,5). The element which contains (4,5) is clearly in $S$.

For order 12, note that the stabiliser of a point in $A_5$ is a version of $A_4$, which has order 12, and there are five such subgroups, and that accounts for all of the maximal subgroups of order 12. Looking at the elements of $S$, the two 3-cycles fix points 4 and 5, while the elements of order 2 fix 3,2,1 respectively (as given in the question) - which gives a non-trivial element in each stabiliser.

Not sure how enlightening this is ...

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As an alternative argument for order 12, observe that $6 \times 12 = 72 > 60$ and, for subgroups $A$ and $B$ of $G$, we have $|AB| = |A||B|/|A \cap B|$. –  Derek Holt Oct 2 '12 at 10:13
Thank you very much. –  user28083 Oct 2 '12 at 17:24