I had a question I was hoping for some help on:

Find all of the subgroups of $A_4$

Here is what I know:

$A_4$ is the alternating group on 4 letters. That is it is the set of all even permutations. The elements are:

$(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)$

which totals to 12 elements. Which means, the subgroups should have order 1,2,3,4,6 and 12. The groups of order 1 and order 12 are trivial. I also know the groups of order 2 which are:

$[1,(12)(34)], [1, (13)(24)], [1, (14)(23)]$

but would someone be able to help me find the subgroups of order 3, 4 and 6 and be able to provide a good, easy explanation of how you got to them for me? I'm pretty sure there is no such groups of order 6, but I don't know how to explain why.

Thank you so much for your help in advance, I really appreciate it!

  • $\begingroup$ A subgroup of order 3 is isomorphic to $\Bbb Z_3$, so it is generated by an element of order 3. To identify the subgroups of order 3 it suffices to identify the elements of order 3. $\endgroup$ – MJD Oct 25 '14 at 16:37
  • $\begingroup$ You mean $(123),(132),(124),(142),(134),(143),(234),(243)$ ? $\endgroup$ – Billy Thorton Oct 25 '14 at 16:38
  • $\begingroup$ Yes, Billy, those are the elements of order 3. $\endgroup$ – Gerry Myerson Oct 27 '14 at 9:46
  • $\begingroup$ Any thoughts about the answers you have received? $\endgroup$ – Gerry Myerson Oct 29 '14 at 3:51
  • 1
    $\begingroup$ I apologize for the late reply - got caught up at school and forgot to revisit. I think most of the answers seem to mask sense - It's actually hard to choose a best answer because both have great qualities! $\endgroup$ – Billy Thorton Nov 10 '14 at 2:49

If you know that the only groups of order 4 (up to isomorphism) are the cyclic group and the Klein 4-group, and that the only groups of order 6 (up to isomorphism) are the cyclic group and $S_3$, then you can just look for copies of those groups in $A_4$.

For a cyclic group of order 4, you need an element of order 4. Are there any in $A_4$?

For a Klein 4-group, you need three elements of order 2, each of which commutes with the other two. Are there such elements in $A_4$?

For a cyclic group of order 6, you need an element of order 6. Are there any in $A_4$?

For a copy of $S_3$, you need an element $a$ of order 2, and an element $b$ of order 3, and you need $ba=ab^2$. Now you have a bit of work to do here, since there are several elements of order 2, and several of order 3; but by the inherent symmetries of the set-up, you can assume the element of order 3 is $(123)$, and then there aren't so many cases you have to look at.

EDIT: but here's an easier way to handle $S_3$. $S_3$ has 3 elements of order 2, none of which commute with the others. Does $A_4$ have that?


If you know Sylow's 1st theorem, then you can easily find the subgroup(s) of order $4$.

By Sylow's 1st theorem, $A_4$ must have at least one subgroup of order $4$. By Lagrange's theorem, the order of every element must divide the order of the group, so the elements of a group of order $4$ can only have orders $1$, $2$ or $4$. Now you can form the Sylow $2$-subgroup(s) by looking at your list of elements!

There are many ways to show that $A_4$ has no subgroup of order $6$. Counting elements and making some multiplications may be the most elementary way.

If $H \leq A_4$ and $|H| = 6$, then $H$ has at least one element of order $2$ and one of order $3$ (by Cauchy's theorem). Let $(a \ b \ c) \in H$. Then also $(a \ b \ c)^2 \in H$. You know that one of the $3$ elements of order $2$ is in $H$. You can check all three individually by multiplying them from both sides with the $3$-cycle and its square, and you should end up with a contradiction in all three cases.

  • $\begingroup$ You can even avoid using Cauchy's theorem: As you need $6$ elements for $H$, it's clear that there must be a $3$-cycle in $H$ (otherwise not enough elements). Now $3$-cycles come in pairs (a $3$-cycle and its square). Thus you have an even number of $3$-cycles together with the neutral element in $H$. Hence you need at least on element of order $2$ to make it to six. $\endgroup$ – Leppala Oct 27 '14 at 9:28

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