Exercise 1704Algebra › GroupsL1Added 3 Nov 2026

The first isomorphism theorem

#first isomorphism theorem#quotient#kernel

Statement

Let $f:G\to G'$ be a group homomorphism. Show that $G/\ker f\cong\operatorname{Im} f$ .

Hint

Define $\bar f:G/\ker f\to\operatorname{Im} f$ by $\bar f(gK)=f(g)$ where $K=\ker f$ . Show it is a well-defined isomorphism.

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Well-defined. If $g_1K=g_2K$ , then $g_2^{-1}g_1\in K$ , so $f(g_2^{-1}g_1)=e'$ , hence $f(g_1)=f(g_2)$ .

Homomorphism. $\bar f(g_1K\cdot g_2K)=\bar f(g_1g_2K)=f(g_1g_2)=f(g_1)f(g_2)=\bar f(g_1K)\bar f(g_2K)$ .

Injective. If $\bar f(gK)=e'$ , then $f(g)=e'$ , so $g\in K$ , hence $gK=K$ .

Surjective. Every element of $\operatorname{Im} f$ is $f(g)=\bar f(gK)$ .

Conclusion. $\bar f$ is an isomorphism, so $G/\ker f\cong\operatorname{Im} f$ .

Consequence. $|G|=|\ker f|\cdot|\operatorname{Im} f|$ for finite groups.