Is the kernel of a group homomorphism unique?

Question

I just began learning group theory and this question came across my mind.

Suppose $f :G \to H$ is a group homomorphism, where $G$ and $H$ are groups. Is $\ker f $ unique?

Based on my understanding, since homomorphism preserves identity from $G$ to $H$, and that identity in groups $G$ and $H$ is unique, there is only a unique element in $\ker f $.

Answer 1

Your argument only shows that the identity element of $G$ is contained in the kernel of $f$. There might be more elements depending on $f$ though.

To give an extreme example: Let $G$ and $H$ be arbitrary groups and consider the homomorphism $1 \colon G \rightarrow H$ of groups that sends each element of $G$ to the identity element of $H$. Then the kernel of $1$ coincides with the group $G$.

Answer 2

By definition$$\ker f=\{g\in G\mid f(g)=e_H\}.\tag1$$So, if you are asking whether $\ker f$ is a unique set, the answer is affirmative, since $\ker f$ is defined by $(1)$; it's a single set. But it seems to me that the question is in fact about whether $\ker f$ consists only of $e_G$. In general, no. If $G$ has more than one element and if $(\forall g\in G):f(g)=e_H$, then $\ker f=G\ne\{e_g\}$.

Answer 3

Kernel of a homomorphism $f:G \to H$ is a set and the set is uniquely defined. If $0$ is the identity in $H$, then the $\ker(f) = \{g\in G \ | \ f(g) = 0\}$. However, the kernel need not have a unique element. Look at the following example:

Let $G$ be the group of integers $\mathbb{Z}$ under addition. Consider the set $\{0,1\}$ and check that the set along with the operation defined as $0+0 = 0 = 1 + 1$ and $0+1 = 1 = 1+0$ forms a group. Call this group $H$. Further check that the operation $f: G \to H$ defined as $f(z) = 0$ if $z$ is even and $f(z) = 1$ if $z$ is odd is a homomorphism. Finally check that the kernel of $f$ is the set of even integers.