As a follow-up from this question: Groups of order 42
I realize that this would probably mean that $G$ is cyclic, could someone walk me through the logic behind this?
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The question is:
If $G$ is a group of order $42$ and we have a surjective homomorphism $\varphi: G \to C_{21}$, must $G$ be cyclic?
We have seen that $\ker\varphi\subseteq Z(G)$. This implies that $G/Z(G)$ is isomorphic to a subgroup of $C_{21}$ and so is cyclic. It is well known that
If $G/Z(G)$ is cyclic, then $G$ is abelian.
Therefore, $G$ is abelian. Since $42=2\cdot3\cdot7$ is a product of distinct primes, $G$ must be cyclic.
(Indeed, there are elements $g_p \in G$ of order $p$ for $p=2,3,7$. Then $g_2g_3$ has order $6$ and $g_2g_3g_7$ has order $42$.)
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And I thought that well-known fact was rather silly... – lhf Nov 14 '15 at 01:47
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Is it always so that if product of the order of a abelian group is composed of distinct primes then it must ne cyclic? – Kushal Bhuyan Nov 14 '15 at 01:52
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@KprimeX, yes, it is. The proof is as in the last sentence. – lhf Nov 14 '15 at 01:53
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Is there any other way to show that it is abelian? – user92821 Nov 14 '15 at 02:13
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I also presume that C21 is <21> right? – user92821 Nov 14 '15 at 02:13
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@user92821, I know no other way to prove that $G$ is abelian. $C_{21}$ is the cyclic group of order $21$. – lhf Nov 14 '15 at 10:17