Kernel of a group homomorphism

Given a Group homomorphism $𝑓 : 𝐺 \to 𝐻$ , with identity $𝑒 𝐻 \in 𝐻$ , the kernel $k e r 𝑓$ is defined as #m/def/group

k e r 𝑓 = {𝑔 \in 𝐺 : 𝑓 (𝑔) = 𝑒 𝐻}

Every kernel is a Normal subgroup, #m/thm/group and vice versa (see quotient group).

Proof of normal subgroup

Clearly $𝑒 \in k e r 𝑓$ . Let $𝑎, 𝑏 \in k e r 𝑓$ , then $𝑓 (𝑎) = 𝑓 (𝑏) = 𝑒$ . It follows $𝑓 (𝑎 𝑏 - 1) = 𝑓 (𝑎) 𝑓 (𝑏 - 1) = 𝑓 (𝑎) 𝑓 (𝑏) - 1 = 𝑒$ , thus $𝑎 𝑏 - 1 \in k e r 𝑓$ . Therefore $k e r 𝑓$ is a subgroup by One step subgroup test. Now let $𝑘 \in k e r 𝑓$ . Then for any $𝑔 \in 𝐺$ , $𝑓 (𝑔 𝑘 𝑔 - 1) = 𝑓 (𝑔) 𝑓 (𝑘) 𝑓 (𝑔 - 1) = 𝑓 (𝑔) 𝑒 𝑓 (𝑔 - 1) = 𝑓 (𝑒) = 𝑒$ , whence $𝑔 𝑘 𝑔 - 1 \in k e r 𝑓$ . Therefore $k e r 𝑓$ is a Normal subgroup.

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