The kernel of a homomorphism is a normal subgroup

The Mathematician — Thu, 15 Sep 2022 13:00:12 +0000

The kernel of a homomorphism is a normal subgroup

Proof. Take

\phi: G \longrightarrow G

and

\begin{align*} \ker(\phi) = \{a\in G \ | \ \phi(a) = e\}, \end{align*}

where

e

is the identity element of

G

. We need to show that for all

g\in G

we have

gag^{-1} \in \ker(\phi)

. Now if

a \in \ker(\phi)

and

g \in G

, then

\begin{align*} \phi(gag^{-1}) &= \phi(g)\phi(a)\phi(g^{-1}) \quad \text{because kernel is a homomorphism} \\ &= \phi(g)\phi(a)\phi(g)^{-1}\\ &= \phi(g)e\phi(g)^{-1} \\ &= \phi(g)\phi(g)^{-1} \\ &= e. \end{align*}

gag^{-1} \in \ker(\phi)

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