Let $G_1$, $G_2$ be groups,$p \neq q$ primes. $|G_1| = 2p$, $|G_2|= 2q$ and $\phi: G_1 \to G_2$ non-trivial homomorphism, find $|\ker(\phi)|$.

Question

Suppose that $G_1$ and $G_2$ are groups, $p$ and $q$ distinct primes, and $|G_1| = 2p$ and $|G_2|= 2q$. Suppose that $\phi: G_1 \to G_2$ is a non-trivial homomorphism. Find $|\ker(\phi)|$.

$\textbf{Proof:}$ We have that $\ker(\phi) \lhd G_1$ (by first isomorphism theorem). By Lagrange's Theorem $|\ker(\phi)|$ divide $|G_1| = 2p$. Then $|\ker(\phi)| \in \{2p,p,2,1\}$

$\hspace{10pt}$Consider the case when $|\ker(\phi)| = 2p$ implies that $\phi(g_1) = e_{G_2}, \forall g_1 \in G_1$ and contradicts the assumption that $\phi$ is a non-trivial homomorphism.

Also, $$ \lvert G_1\rvert = \lvert \ker\phi\rvert\cdot \lvert \mbox{im}\phi\rvert $$ Hence, $$ 2p = \lvert \ker\phi\rvert \cdot \gcd(\lvert G_1\rvert, \lvert G_2\rvert) = \lvert \ker\phi\rvert \cdot \gcd(2p,2q) = \lvert \ker\phi\rvert \cdot 2 $$

$$ \therefore \lvert \mbox{im}\phi\rvert = 2, \hspace{10pt} \therefore\lvert \ker\phi\rvert = p. $$

Answer 1

Hint: $|im(\phi)|$ divides both $|G_1|$ and $|G_2|$.

... and so $|im(\phi)|$ divides their gcd, which is $2$. So, $|im(\phi)|=1$ or $|im(\phi)|=2$. Since $\phi$ is not trivial, $|im(\phi)|=2$. Therefore, $|\ker(\phi)|=p$.