# Why $A_{5}$ has no subgroup of order $15$?

Please show that $A_{5}$, a group of order $60$, has no subgroup of order $15$.

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I know A_{5} is a simple group – Ali Apr 23 '12 at 5:58
For now, I'll leave this answer with just a hint. You may want to fill in the gaps... If this is homework, please tag it as homework. – user21436 Apr 23 '12 at 6:00
@Ali That's all you know about $A_5$? Do you not know what the elements are? What the elements of order 3 and 5 are? – Alex B. Apr 23 '12 at 6:01

Show that every group of order $15$ is cyclic. The result follows since there is no element of order $15$ in $A_5$.

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Hint

• $A_5$ is simple.

• What is the index of such a group? Let $A_5$, a simple group act on left cosets of this proper subgroup? What can you say about the kernel of the homomorphism that comes with this action?

• So, now apply first isomorphism theorem; Lagrange's theorem to conclude a result known due to Poincare...

• So, what do you conclude?

Perhaps, a more adhoc solution that applies exclusively here, but nonetheless, an important fact would be to prove the following:

• $A_5$ has no element of order $15$. (Perhaps, you should try to list all those orders that occur in $A_5$.)

• A group of order $15$ is cyclic. (Perhaps, I suggest you classify groups of order $pq$ for primes $p$ and $q$. This is a fun exercise and I suggest you'll do this. You'll get comfortable thinking about group actions and Sylow's theorem. )

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Please explain more – Ali Apr 23 '12 at 6:00
there is a homomorphism: $G$ to sym(4). Hence $G$ has a non trival normal subgroup that is contradiction – Ali Apr 23 '12 at 6:17
@Ali You're uttering the right words in wrong order. Perhaps, try a bit more harder. You're closer to the answer. Think a bit more before I'll tell you anything, Regards, – user21436 Apr 23 '12 at 6:22
thank you very much – Ali Apr 23 '12 at 6:23
I don't think the second solution is ad hoc, the exact same argument works when $A_5$ is replaced by $S_n$ or $A_n$ for $n \leq 7$. – Mikko Korhonen Apr 23 '12 at 6:29

Assume $H \leq A_5$ with $|H| = 15$ and let $X:=\{gH \mid g \in G\}$. Then $\# X = 4$. $G$ acts op $X$ by left multiplication i.e. $g'(gH) = (g'g)H$. Let $\alpha \in A_5$ be a 5-cycle. Then $\langle \alpha\rangle$ does act on $X$,too. But the length of an orbit divides the group-order which is 5. But $\# X = 4 < 5$ so each orbit contains only one element. That means $\alpha H = H$ for all $\alpha$. So $\alpha \in H$. There are 24 of those $\alpha$. Contradition because $H$ cannot contain more than 15 elements.

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