# Group Homomorphisms From Group of Order 21 to Group of Order 49

## Problem 346

Let $G$ be a finite group of order $21$ and let $K$ be a finite group of order $49$.
Suppose that $G$ does not have a normal subgroup of order $3$.
Then determine all group homomorphisms from $G$ to $K$.

## Proof.

Let $e$ be the identity element of the group $K$.
We claim that every group homomorphism from $G$ to $K$ is trivial.
Namely, if $\phi:G \to K$ is a group homomorphism, then we have $\phi(g)=e$ for every $g\in G$.

The first isomorphism theorem gives the isomorphism
$G/\ker(\phi)\cong \im(\phi) < K.$ It follows that the order $|\im(\phi)|$ of the image $\im(\phi)$ is a divisor of the order of $G$ and that of $K$. Hence the order $|\im(\phi)|$ divides the greatest common divisor of $|G|=3\cdot 7$ and $|K|=7^2$, which is $7$. So, the possibilities are $|\im(\phi)|=1, 7$. If $|\im(\phi)|=7$, then we have $\frac{|G|}{|\ker(\phi)|}=|\im(\phi)|=7,$ and we obtain $|\ker(\phi)|=3$. Since the kernel of a group homomorphism is a normal subgroup, this contradicts the assumption that $G$ does not have a normal subgroup of order $3$. Therefore, we must have $|\im(\phi)|=1$, and this implies that $\phi$ is a trivial homomorphism. Thus we conclude that every group homomorphism from $G$ to $K$ is trivial.

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