Groups of order 42

Question

Drawing from another question that was recently posted:

If $G$ is a group of order 42 and we have $\varphi:G\to G$ with $\varphi(G)$ isomorphic to $\mathbb{Z}/21\mathbb{Z}$.

How can we show that the kernel of $\varphi$ is a subset of $Z(G)$?

I thought of using the First Isomorphism Theorem, but that would only help us to show that the kernel is a normal subgroup of $G$. Any thoughts?

score 5 · Answer 1 · answered Nov 14 '15 at 00:18

5

$K=\ker\varphi$ has order $2$ and so is cyclic. Let $k$ be a generator of $K$.

Take $g\in G$. Then $gK=Kg$ and so $gk=k'g$ with $k'\in K=\{e,k\}$. Now $k'$ must be $k$ because $k\ne e$. This proves that $k$ commutes with all elements of $G$ and so $K\subseteq Z(G)$.

answered Nov 14 '15 at 00:18

lhf

216,483

So does this meant that G is abelian and cyclic? – user115017 Nov 14 '15 at 00:47
Yes, see http://math.stackexchange.com/questions/1527982/group-of-order-42-cyclic. – lhf Nov 14 '15 at 01:45

Groups of order 42

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