Common Findings of Homomorphisms

Theorem

Let $G, H$ be Groups and let $φ : G \to H$ be a Homomorphism (or Group Morphism). Then:

$φ (1_{G}) = 1_{H}$ where $1_{G}, 1_{H}$ are the identities of $G, H$ respectively.
$φ (g^{- 1}) = φ (g)^{- 1}$
$φ (g^{n}) = φ (g)^{n}$ for all $n \in Z$
$Ker (φ)$ is a Subgroup of $G$ . (see Kernel)
$im (φ)$ , the image of $G$ under $φ$ , is a subgroup of $H$ .

Proof

$φ (1_{G}) = φ (1_{G} 1_{G}) = φ (1_{G}) φ (1_{G})$ so then using Cancellation Theorem then (1) holds.
$φ (1_{G}) = φ (g g^{- 1}) = φ (g) φ (g^{- 1})$ and using (1) then this equals $1_{H}$ . Hence:

1_{H} = φ (g) φ (g^{- 1})

Multiplying both sides on the left by $φ (g)$ and simplifying gives (2).

See 16 Homomorphisms, Isomorphisms#1.
Since $1_{G} \in Ker (φ)$ then the kernel is not empty. Let $x, y \in Ker (φ)$ , that is $φ (x) = φ (y) = 1_{H}$ . Then:

φ (x y^{- 1}) = φ (x) φ (y^{- 1}) = φ (x) φ (y)^{- 1} = 1_{H} 1_{H}^{- 1} = 1_{H}

that is $x y^{- 1} \in Ker (φ)$ . By the Subgroup#The Subgroup Criterion (combining both), then $Ker (φ) \leq G$ .

Since $φ (1_{G}) = 1_{H}$ then the identity of $H$ lies in the image of $φ$ , so $im (φ)$ is nonempty. If $x, y$ are in $im (φ)$ , say $x = φ (a), y = φ (b)$ , then $y^{- 1} = φ (b^{- 1})$ then by (2) then $x y^{- 1} = φ (a) φ (b^{- 1}) = φ (a b^{- 1})$ as $φ$ is a homomorphism. Now since $x y^{- 1} \in im (φ)$ then it is a subgroup of $H$ by the subgroup criterion.

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