Does A5 have a subgroup of order 6?

I am trying to figure out if $A_5$ has a subgroup of order $6$. Rather than a yes/no answer I would prefer if someone could show me how they find their way to the answer. Below is my false attempt at a solution. Ideally I could get some tips on how to fix my attempt, but if it seems a dead end alternate solutions are really appreciated:).

I began by counting the number of elements of each cycle type in S5 so that I could deduce the conjugacy classes and class equation for S5. Next I checked centralizers to determine which were conjugacy classes in A5 and I got the class equation for $A_5$ as $60 = 1 + 12 + 12 + 15 + 20$.

Using the orbit-stabilizer theorem, we can deduce from the order of the last conjugacy class that there is a subgroup of order $3$, namely the centralizer of any element in this conjugacy class, for example $C((12345))={e, (12345), (15432)}$. It is easy to then find a subgroup of order $2$ such as $<(12)(34)>= {e, (12)(34)}$. Since the intersection of these two groups is ${e}$, my hope was that $<(12)(34)>(12345)$ would be a subgroup of order $6$, unfortunately not all the elements commute (If the elements of two subgroups $H$,$K$ commute and their intersection is ${e}$ then $HK$ is a subgroup with order equal to the product of the orders) so I'm left at a dead end.

• $(1,2,3,4,5)$ is an element of order $5$, so it couldn't possibly be contained in a subgroup of order $6$. And your final sentence does not make much sense. Why would you expect all elements to commute? The subgroup you are looking for might be a nonabelian subgroup of order $6$. – Derek Holt Apr 22 '15 at 14:13
• Ah yes, that make sense. So this is definitely a dead end. If the elements of two subgroups H,K commute and their intersection is {e} then HK is a subgroup with order equal to the product of the orders. – mathguy Apr 22 '15 at 14:16
• If you want to cheat, you can look at Twisted S3 in A5 and the pages it links at the top of the page. – Jeppe Stig Nielsen Apr 22 '15 at 14:56
• If you are able to calculate the normalizer of $\langle(123)\rangle$, then you have just to check if it contains an element of order $2$. – j.p. Apr 22 '15 at 15:26

If $\pi\in S_5$ is a permutaion of $\{1,2,3\}$, then either $\pi$ or $\pi\circ (4\,5)$ is an even permutation, i.e., $\in A_5$. This allows us to embed $S_3\to A_5$.
Or have fun with geometry: $A_5$ is the symmetry group of the dodekahedron. It is possible to select $4$ of its $20$ vertices that make up a regular tetrahedron $T_1$. A rotation of the dodekahedron by $72^\circ$ turns $T_1$ into another such tetrahedron $T_2$. Then the subgroup of $A_5$ that fixes $T_1\cup T_2$ turns out to be of isomorphic to $S_3$ (why?)
Although you said that you didn't want a yes/no answer, my hint to you is that the answer is yes, so you should just try to construct a subgroup, $H$, of order 6. Look for elements $\sigma$ and $\rho$ of order 2 or 3. A priori there are two possibilities for $H$ - namely $\mathbb Z_6=\langle\rho, \sigma | \sigma \rho \sigma ^{-1}=\rho, \quad \rho^3=e=\sigma^2\rangle$ and $D_6=\langle\rho, \sigma | \sigma \rho \sigma ^{-1}=\rho^{-1}, \quad \rho^3=e=\sigma^2\rangle.$ $\mathbb Z_6$ has an element of order 6, but no such element exists in $A_5$ - you'd need a product of 2-cycles and a 3-cycles, but since you're only allowed to permute 5 objects, you can only get elements like $(12)(345)$ which are odd, so not in $A_5$. So the only possibility is $D_6$, but if you choose your elements $\sigma$ and $\rho$ well you'll get this.
• It is not true that every element of order $2$ together with every element of order $3$ generates a subgroup of order $6$, so you may or may not be home, depending on your choice. – Derek Holt Apr 22 '15 at 14:38