A homomorphism is injective iff its kernel is trivial

Remarks

If $\varphi(a) = \varphi(b)$, then $\varphi(a^{-1}b) = e_H$, so $a^{-1}b \in \ker \varphi$. More generally, $\varphi(a) = \varphi(b)$ exactly when $a$ and $b$ lie in the same coset of $\ker \varphi$.

Going deeper

For a student

For a map of sets, injectivity needs a check on each pair of points. For a homomorphism, it is enough to look at one fibre, the kernel.

The proof uses the group structure to move any pair to the identity:

1. If $\varphi$ is injective, then only $e$ maps to $e$, because $\varphi(e) = e$.
2. Conversely, suppose that $\ker \varphi = \{e\}$ and $\varphi(a) = \varphi(b)$.
3. Then $\varphi(a^{-1}b) = \varphi(a)^{-1}\varphi(b) = e$.
4. Thus $a^{-1}b = e$, and $a = b$.

A worked example with $\varphi : \mathbb{Z} \to \mathbb{Z}$, $\varphi(n) = 3n$:

1. The map is a homomorphism, because $3(m + n) = 3m + 3n$.
2. If $3n = 0$, then $n = 0$. Thus the kernel is $\{0\}$.
3. By the result, $\varphi$ is injective. There is no need to compare pairs.

A non-example is reduction modulo $3$, $\mathbb{Z} \to \mathbb{Z}/3\mathbb{Z}$. Its kernel is $3\mathbb{Z}$, which is not trivial. Thus the map is not injective: for example, $0$ and $3$ have the same image.

More generally, $\varphi(a) = \varphi(b)$ exactly when $a^{-1}b \in \ker\varphi$, that is, when $a$ and $b$ are in the same coset of the kernel. Thus each fibre of $\varphi$ is a coset of the kernel, or it is empty.

A common error is to use the test for a map that is not a homomorphism. The map $x \mapsto x^2$ on $\mathbb{R}$ sends only $0$ to $0$, but it is not injective.