# Alternating group of 5 elements and its generators.

Show $A_5 = (\langle 1,2,3\rangle, \langle1,2,3,4,5\rangle)$. Guys how can i show this in the best way .

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First, you have to know what's in $A_5$. Then, you can show that various clever products of the alleged generators give you all those elements. Do you know how to prove $(1,2)$ and $(1,2,\dots,n)$ generate $S_n$? It's a similar sort of argument. –  Gerry Myerson Apr 14 '12 at 12:42
Well they certainly generate a subgroup of $A_5$ and since the elements you are given have orders 3, and 5 you can tell something about the order of the subgroup, and therefore its index in $A_5$. How you go from there depends what knowledge you are allowed to use. If you are allowed to know about $A_5$ then you know what subgroups it has. Or you could investigate the order of the stabiliser of a point. Or see if your subgroup has any elements of even order. Or use the fact that groups of certain orders have to be abelian. –  Mark Bennet Apr 14 '12 at 12:45

Spoiler alert. Call the generating elements $p$ of order 3 and $q$ of order 5. We know that the order of $A_5$ is 60 and we check that $pq \neq qp$. If we are allowed some basic knowledge about groups, that is the only calculation we have to do.

The order of the subgroup generated by $p$ and $q$ must be a multiple of 15, since the orders of the elements must be factors of the orders of the group. It can't be 15, because a group of order 15 would be abelian. The only possibilities are therefore 30 and 60. Now a subgroup of order 30 would have index 2, and be thus be normal. But $A_5$ is simple, so this cannot happen. So we must have the whole group.

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At the opposite end of the spectrum from Auke's answer, here's a solution involving no calculations, beyond the fact that the given elements have order $3$ and $5$. All you need is that $A_5$ is simple of order $60$ and some general group theory.

If $A_5$ has a subgroup $H$ of index $i>1$, then $A_5$ acts nontrivially on the left cosets of $H$ by left multiplication. Since $A_5$ is simple, this action must be faithful, and so $A_5$ embeds in $S_i$. Thus we must have $|A_5| \le |S_i| =i!$, and so $i \ge 5$. The given subgroup $(\langle 1,2,3\rangle, \langle1,2,3,4,5\rangle)$ has order divisible by $3$ and $5$, and hence at least $15$. Thus the index is at most $4$, and so must be $1$: i.e., the subgroup is all of $A_5$.

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I like that - it goes with the representation theory tag (assuming that it is referring to representing groups as permutation groups) –  Mark Bennet Apr 14 '12 at 13:14
@Mark I took off the representation theory tag because I thought that it was for questions about matrix representations. The tag's description seems to confirm this. –  MJD Apr 14 '12 at 13:56
@MarkDominus - I was too lazy to check! –  Mark Bennet Apr 14 '12 at 14:03

If you want a different kind of answer with more calculation in it, note that the subgroup generated by $p$ (order 3) and $q$ (order 5) acts transitively on the underlying set (the 5-cycle does that) so the stabiliser of a point will have index 5 within the subgroup generated by $p$ and $q$ - and this stabiliser is a subgroup within $A_5$, so its order will be a factor of $60/5 = 12$.

$qpq^{-1}$ is another 3-cycle (conjugation preserves cycle-type). And it will be found that the two 3-cycles both fix one of the underlying points. A few calculations later and it will be seen that the stabiliser has order greater than 6, so it must be 12.

It follows that the whole subgroup has order 60, as required.

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Well, "best" is of course a bit vague, but one of the most straightforward ways is by showing that you can combine these generators into the normal generators of $A_5$, ie. all the 2,2-cycles in $A_5$, making $A_5$ a subgroup of your group, and then show that your group is a subgroup of $A_5$ (which is obvious), making them equal. There are quite a number of 2,2-cycles in $A_5$, however, so it's not a quick method.

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