Prove $\ker(\phi)\ne \{e\}$

Question

$G_1$ and $G_2$ are finite groups such that $|G_1|>|G_2|$ and $|G_2|$ divides $|G_1|$.
Let $\phi : G_1\rightarrow G_2\space$ be a homomorphism such that $\ker(\phi)\ne G_1$.

Can I conclude that $\ker(\phi)\ne \{e\}$?
It makes sense but I can't prove it.

Answer 1

$\frac{G}{\ker\varphi}\cong \varphi(G)$. Notice $\varphi(G)$ is a subgroup of $H$ so $|\varphi(G)|\leq|H|<|G|$.

Since isomorphism are bijections $|G|=|\ker\varphi||\varphi(G)|$. Since $|\varphi(G)|<|G|$ conclude $|\ker\varphi|>1$.

Notice you don't need divisibility.

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Here is a shorter wording, a homomorphism is injective if and only if it has a trivial kernel. However an injective function from $G$ to $H$ if $|G|>|H|$ is impossible by definition.

Answer 2

Since you know that a homomorphism $ \phi : G_1 \to G_2 $ is injective $\iff \ker(\phi) = \{e\} $ you can proof by contradiction.

If we assume $ \ker(\phi) = \{e\} $, it follows that $\phi$ is injective. This is impossible because of $|G_1| \gt |G_2|$ and both groups are finite.

Answer 3

If $\ker \phi=\{e\}$, $\phi$ is injective and $G_1$ is thus isomorphic to a subgroup of $G_2$, hence $\lvert G_1\rvert \le\lvert G_2\rvert $, which contradicts $\lvert G_1\rvert >\lvert G_2\rvert $. The only required hypothesis is the latter.