2

Let $G$ be a non-abelian group of order $6$ with exactly three elements of order $2$. Show that the conjugation action on the set of elements of order $2$ induces an isomorphism.

I just need to show that the kernel of the action is trivial. Not sure how to go about doing that. I think maybe a proof by contradiction but I can't find a contradiction. I would think it would violate "non-abelian-ness" of the group. Thanks for any help!

Brian Fitzpatrick
  • 26,933
Rebekah
  • 1,037

3 Answers3

2

If you Sylow Theorems, you get 1 Sylow 3-subgroup and either 1 or 3 Sylow 2-subgroup(s).

If G has 1 Sylow 2-subgroup, it must be normal since it is a unique subgroup of given order. So we can take direct product of these two subgroups, which is isomorphic to G. But since each Sylow subgroups here are isomorphic to $\mathbb{Z}_2$ and $\mathbb{Z}_3$, we have that $G \cong \mathbb{Z}_2 \times \mathbb{Z}_3$, which contradicts that $G$ is not abelian.

So G has 3 Sylow 2-subgroups. Now you can explicitly list out elements of $G$ to see why it is isomorphic to $S_3$

JKim
  • 101
  • Could you explain how we can list out the elements of $G$? The three Sylow-2 subgroups give us three elements of order 2, and we know the Sylow-2 groups are conjugates. However, I'm not sure how to set up the entire multiplication table, as I seem to be missing some information. – Sha Vuklia May 04 '19 at 11:33
2

Here is a hands-on method.

Note that $G$ has an element $a$ of order $3$ hence at least two as $a^2$ has order $3$, but can't have an element of order $6$ or it would be cyclic and hence abelian.

Suppose the elements of order $2$ are $b,c,d$, then the elements of the group are $1,a,a^2,b,c,d$. No element of order $3$ can commute with any element of order $2$ else the product would have order $6$

Now $ab\neq ba$ implies both $aba^{-1} \neq b$ and $bab^{-1} \neq a$ - so neither the elements of order $3$ nor those of order $2$ can have a trivial action.

Mark Bennet
  • 101,769
  • Just an addendum, this doesn't motivate why $G$ must have an element of order 3. It's important to note that if none of $G$'s elements have order 3 then (by lagranges theorem) they must have order dividing 6, that isn't 6 or 3 (and only the identity has 1) and thus must all be order 2. That group however is $Z_2 \times Z_2 \times Z_2$ which is abelian (although I haven't proven why this MUST be the case) – Sidharth Ghoshal Nov 15 '15 at 22:42
  • 2
    @frogeyedpeas Alternatively, count the elements. If there us an element of order $6$ the group is cyclic, and that doesn't work. By Lagrange, the possible orders are $1,2,3$ . There is one element of order $1$ and three of order $2$ (given) so there must be (two) elements of order three to get up to six elements. – Mark Bennet Nov 15 '15 at 22:46
  • @Mark Bennet Sir I do not understand $6^{th}$ line where you had mentioned $ab\neq ba$ implies both $aba^{-1}\neq a $ and other .How Sir? – Curious student May 01 '18 at 16:56
  • @SRJ If $ab\neq ba$ then $(ab)a^{-1}\neq (ba)a^{-1}$ because you have post-multiplied both sides of an equality by the same thing, then use the associative property. You can pre-multiply by $b^{-1}$ to get the other inequality. – Mark Bennet May 01 '18 at 20:53
  • 1
    @MarkBennet: That would give $aba^{-1} \neq b$ instead of $\neq a$ (and similarly for the other case). – Aryaman Maithani Jun 12 '21 at 12:59
  • @AryamanMaithani Good spot - have corrected – Mark Bennet Jun 12 '21 at 13:02
0

Hint: Suppose $x\in G$ is an element of the kernel of the action, i.e. fixes the three involutions under conjugation. What do you know about the group generated by the three involution, and what does that tell you about $x$?

jpvee
  • 3,643
  • 2
  • 24
  • 33