11
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $(1,2,3,4,5)$ 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 $6$. $\endgroup$ – Derek Holt Apr 22 '15 at 14:13
  • $\begingroup$ 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. $\endgroup$ – mathguy Apr 22 '15 at 14:16
  • $\begingroup$ If you want to cheat, you can look at Twisted S3 in A5 and the pages it links at the top of the page. $\endgroup$ – Jeppe Stig Nielsen Apr 22 '15 at 14:56
  • 1
    $\begingroup$ 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 $2$. $\endgroup$ – j.p. Apr 22 '15 at 15:26
12
$\begingroup$

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?)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You're only "home" if the two elements commute. $\endgroup$ – Matt Samuel Apr 22 '15 at 14:32
  • $\begingroup$ I don't think they will commute. But I'm pretty sure you're still home. $\endgroup$ – James Apr 22 '15 at 14:37
  • 2
    $\begingroup$ 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. $\endgroup$ – Derek Holt Apr 22 '15 at 14:38
  • $\begingroup$ Good point. I'll edit my answer. $\endgroup$ – James Apr 22 '15 at 14:43
  • 2
    $\begingroup$ You should just try and find the elements - it's not so hard! $\endgroup$ – Derek Holt Apr 22 '15 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.