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

Question

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

Answer 1

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

Answer 2

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?

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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. )

Answer 3

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.