Exercise 977Algebra › GroupsL1Added 3 Nov 2026

Injectivity and kernel

#kernel#injectivity#homomorphism

Statement

Let $f:G\to G'$ be a group homomorphism. Show that $f$ is injective iff $\ker f=\{e\}$ .

Hint

For the forward direction, use $f(e)=e'$ . For the converse, assume $f(a)=f(b)$ .

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Solution

If injective. If $g\in\ker f$ , then $f(g)=e'=f(e)$ , so $g=e$ by injectivity. Hence $\ker f=\{e\}$ .

Conversely. Suppose $\ker f=\{e\}$ . If $f(a)=f(b)$ , then $f(ab^{-1})=f(a)f(b)^{-1}=e'$ , so $ab^{-1}\in\ker f=\{e\}$ , i.e. $a=b$ . So $f$ is injective.

Conclusion. $f$ is injective iff its kernel is trivial.

Injectivity and kernel

Statement

Solution

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