# Smallest $m \mid 60$ so that there is no subgroup $H \leq A_5$ with order $m$

What is the smallest $m \mid 60$ so that there is no subgroup $H \leq A_5$ with order $m$ ?

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i thought that it might be 30 but even if it would be the case i have no idea how to proof that this cant be a subgroup. I already did it for $A_4$ and there it was $6$. – Epsilon Oct 6 '12 at 12:40
Echoing @ChrisEagle's comment, have you tried listing the subgroups of $A_5$? – Olivier Bégassat Oct 6 '12 at 12:41
Or indeed the factors of $60$? – Chris Eagle Oct 6 '12 at 12:44
if i have shown that it is 6 for $A_4$ , is this then also true for $A_5$ ? – Epsilon Oct 6 '12 at 12:45
because if $H \leq A_4$ and $|H| = 6$ then $H$ cant be a subgroup eventough $6 \mid 12$. But if $H \nleq A_4 \leq A_5$ then $H \nleq A_5$ ? – Epsilon Oct 6 '12 at 12:47

The factors of $60$, in order, are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$.

It's easy to see that $A_5$ has cyclic subgroups orders $1, 2, 3$ and $5$, and $\langle (12)(34),(13)(24)\rangle$ is a subgroup of order $4$.

For $m=6$ or $10$, the possible groups of order $m$ are cyclic and dihedral. $A_5$ has no element of either order, so we need to look for dihedral subgroups. Thus we seek elements of order $3$ and $5$ which are conjugated into their inverses by appropriate elements of order $2$. These are easily found: $\langle (123), (12)(45)\rangle$ has order $6$ and $\langle (12345),(14)(23)\rangle$ has order $10$.

The five copies of $A_4$ each have order $12$.

Every group of order $15$ is cyclic, and (by considering possible cycle types) $A_5$ has no element of order $15$, so $A_5$ has no subgroup of order $15$, and $15$ is the smallest such $m$.

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is it for all groups of order 15 true that they are cyclic ? – Epsilon Oct 6 '12 at 13:18
Yes, that's why I say in my answer "every group of order $15$ is cyclic". – Chris Eagle Oct 6 '12 at 13:20
How can you proof that ? – Epsilon Oct 7 '12 at 13:07
The easy way is by Sylow's theorems: the $3$- and $5$-Sylows are both normal, hence the group is their direct product, hence it is $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z} \cong \mathbb{Z}/15\mathbb{Z}$. – Chris Eagle Oct 7 '12 at 22:51
is every group with 15 elements isomorph to $\mathbb Z / 15 \mathbb Z$ ? – Epsilon Oct 8 '12 at 15:40

First,Using Sylow's theorem,we can rule out $2,3,5$

Now considering $D_{10}$ (Dihedral group),we can rule out $10$

Considering $A_4$ , we can see that $m \geq 12$.

Now the next possible value is $15$ ,please check the following thread Why $A_{5}$ has no subgroup of order $15$?

Not sure it is the best method though,there might be some way to find it without explicitly checking for every value of $m$.

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why cant we have $m = 6$ ? – Epsilon Oct 6 '12 at 12:50
Trying to find $S_3$ in $A_5$ by simply permuting $\{1,2,3\}$ fails. But we can fix parity problems by adding the cycle $(4 5)$ as correction if necessary. Thus in detail: $(1 2 3)$ and $(1 2)(4 5)$ generate a subgroup of $A_5$ of order 6. – Hagen von Eitzen Oct 6 '12 at 12:59
And what about $m=5$ or $10$ for that matter? Neither possibility is resolved by considering $A_4$. – Chris Eagle Oct 6 '12 at 13:00
I used $A_4$ as he seemed to have done the cases of $5$ and $10$,I edited my answer . – TheJoker Oct 6 '12 at 13:05
$A_5$ is the symmetry group of the pentagondodekahedron and as such contains $C_5$ and $D_5$ (the stabilizer of a face, the stabilizer of two opposite faces). – Hagen von Eitzen Oct 6 '12 at 13:05