3

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?

user92821
  • 133
  • 7

1 Answers1

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 $K\subseteq Z(G)$.

lhf
  • 216,483