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Let $G$ be a finite group. Show that $[G:Z(G)]$ cannot be prime.

Assume $G$ is abelian. The $Z(G) = G$ and $[G:Z(G)] = 1$, not a prime.

Now consider the case where $G$ is not abelian. If $\frac{G}{Z(G)} = p$ some prime $p$, then this quotient group would be cyclic and the $G$ would be abelian, a contradiction $\#$. Thus, $\frac{G}{Z(G)} = [G:Z(G)] \neq p$ cannot be a prime.

Is it really this simple?

All About Groups
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    No, of course not: how exactly from the quotient group being cyclic follows that $;G;$ is abelian? That's precisely the gist of this! – Timbuc Apr 07 '15 at 19:39
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    Note that $H, G/H$ can both be abelian even for nonabelian $G$, such as $H= A_3, G = S_3$. – Rolf Hoyer Apr 07 '15 at 19:45
  • Timbuc, I have shown by proof that it follows from the quotient group being cyclic that $G$ is abelian. – All About Groups Apr 07 '15 at 19:49
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    @AllAboutGroups No, you haven't at all shown (and by this I mean "proved") that . You just wrote it! You seem to believe that if $;G;$ is not abelian then it can not have cyclic quotients, which of course is false. – Timbuc Apr 07 '15 at 19:51
  • I meant that I had shown below the proof required. – All About Groups Apr 07 '15 at 19:54
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    @AllAboutGroups Yes, below you did, not in your question when you ask "Is it really this simple?"...as your proof shows, it really isn't that simple... – Timbuc Apr 07 '15 at 19:57

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If $\frac{G}{Z(G)}$ is cyclic then there exists some generator $gZ(G)$ such that $\langle gZ(g)\rangle= \frac{G}{Z(G)}$. Let $a, b$ be arbitrary elements of $G$. $aZ(G)$ = $(gZ(G))^i = g^iZ(G)$ for some $i$, and $bZ(G) = (gZ(G))^j = g^jZ(G)$ for some $j$. $aZ(G) = g^iZ(G)$ implies $a = g^iz_1$ for some $z_1 \in Z(G)$. Similarly, $bZ(G) = g^jZ(G)$ implies $b = g^jz_2$ for some $z_2 \in Z(G)$. Thus $ab = (g^iz_1)(g^jz_2) = g^ig^jz_1z_2$ since elements of the center commute with all elements of $G$. We have $ab = g^ig^jz_1z_2 = g^{i+j}z_1z_2 = g^{j+i}z_2z_1 = g^jg^iz_2z_1 = g^jz_2g^iz_1 = ba$. Since $a$ and $b$ were arbitrary, we’ve shown the group operation on $G$ is commutative. Thus $G$ is Abelian.

janmarqz
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