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I'm still reading Gallian's "Contemporary Abstract Algebra".

This is Exercise 7.56 ibid in the section called "Cosets and Lagrange's Theorem," in which, for example, the orbit-stabiliser theorem is proven. Answers that use only the tools available in the textbook prior to the exercise are preferred. Feel free to establish lemmas from them.

The Question:

Why does the fact that $\nexists H\le A_4$ with $\lvert H\rvert=6$ imply $\lvert Z(A_4)\rvert=1$?

Here $A_4$ is defined in terms of even permutations within $S_4$ and the centre of $G$ is $Z(G)$.

Thoughts:

What I'm asked to prove is a property of $A_4$ stronger than just being nonabelian, since, of course, if $\lvert Z(A_4)\rvert=1$, then $A_4$ is nonabelian.

Quotient groups aren't covered yet, so I can't use an approach similar to this answer to this lemma (that I cannot use in the spirit of the question at hand):

Let $G$ be a non-Abelian group, $\lvert{G}\rvert=6$. Then $Z(G)=1$.

So yeah, I don't understand what's going on exactly; I'm sorry I haven't got more to say.

Please help :)

Shaun
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    The point, I think, is to take the centre and form other subgroups. For example, if $Z$ has order $2$, then you can take a subgroup of order $3$ and form their product which must have order $6$. Similarly, if $Z$ has order $3$ (take a subgroup of order $2$). What other possibilities are there? $Z$ does not have order $6$, so $Z$ must have order $4$, but then take a subgroup of $Z$ of order $2$ and proceed as before. – the_fox Feb 23 '19 at 19:32
  • I think that does the trick, since then all one is left with is elements of order two, none of which commute with all $\sigma\in A_4$ (by inspection). – Shaun Feb 23 '19 at 21:33

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