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Let f : G → $Z_2$ be a central extension of $Z_4$ by $Z_2$. Then G is isomorphic to either $Z_8$ or $Z_4$ × $Z_2$.

Proof Since $Z_2$ is cyclic, Lemma 4.7.15 shows that G is abelian. So we could just appeal to the Fundamental Theorem of Finite Abelian Groups, which tells us that $Z_8$, $Z_4$ × $Z_2$, and $Z_2$ × $Z_2$ × $Z_2$ are the only abelian groups of order 8. Clearly, the first two are the only ones that are extensions of Z4.

Alternatively, we can argue directly. Write $Z_4 = <b>$ and $Z_2 = <a>$ and let f(x) = a. Then $f(x^2) = e$, and hence $x^2$ ∈ b. Now x has even order, so that $o(x) = 2o(x^2)$. Thus, if $x^2 = b$ or $b^{−1}$, then x has order 8, and hence G is cyclic.

I don't understand how G is cyclic. If G had order 8 then of course it is...but they don't tell us that G has order 8. So how can we assume that?

If x has order 2, then s(a) = x gives a splitting of f, and hence G is the semidirect product obtained from the action of $Z_2$ on $Z_4$. But in this case, the action is trivial, so that the semidirect product thus obtained is just the direct product.

If x has order 4, then $x^2 = b^2$. But then $b^{−1}x$ has order 2, and setting s(a) = $b^{−1}x$ gives a section of f, and hence $G \cong Z_4 × Z_2$ as above.

I have a question about these two...there is a corollary in the textbook that states, "if $f: G \rightarrow K$ is a split surjection and if G is abelian, then $G \cong kerf$ × $K$. So, using this corollary, don't we automatically get $G \cong Z_4 × Z_2$ without even considering s(a)=x and $s(a)=b^{-1}x$?

Thanks in advance

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For the second part: You need the surjection to be split, which it need not be in all cases. For the first part: This follows directly from the definition of a central extension and the usual isomorphism theorems. – Tobias Kildetoft Mar 20 '13 at 14:11

1 Answer 1

up vote 1 down vote accepted
  1. Lemma 4.7.15 is probably that when $G/Z(G)$ is cyclic, $G$ must be abelian. Since you are making a central extension of $\mathbb{Z}_4$ by $\mathbb{Z}_2$, you are making a group $G$ for which $K\cong \mathbb{Z}_4$ for some $K\leqslant Z(G)$ and $G/K\cong \mathbb{Z}_2$. So Lemma 4.7.15 applies.

  2. Yes, as long as you know that $G$ is a semidirect product (split extension).

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Lemma 4.7.15: Let f : G → $Z_k$ be a central extension of the abelian group H by the cyclic group $Z_k$. Then G is abelian. – user58289 Mar 20 '13 at 14:47
I'm not sure if I understand your first answer. What is "K"? Where did it come from? – user58289 Mar 20 '13 at 14:52
@Artus It is from the definition of a central extension. – Tobias Kildetoft Mar 20 '13 at 15:09
@Alexander: I think you mean $G/K \cong \mathbb{Z}_2$, not $G/Z(G)$.. – spin Mar 20 '13 at 17:08

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