Isomorphisms

Group Isomorphism

An isomorphism from a group

(G, ⋆)

to a group

(H, \cdot)

is a bijective function

ϕ : G \to H

such that for any

a, b \in G

ϕ (a ⋆ b) = ϕ (a) \cdot ϕ (b)

Isomorphic Groups

If there is an isomorphism from

G

H

we say that

G

and

H

are isomorphic and write

G ≅ H

Identities are Carried

Let

e_{G}

be the identity of

G

and

e_{H}

H

, then for any isomorphism from

G

H

, we have

ϕ (e_{G}) = e_{H}

We know $e_{G} = e_{G} e_{G}$ , therefore $ϕ (e_{H}) = ϕ (e_{H} e_{H}) = ϕ (e_{H}) ϕ (e_{H})$ , but also $ϕ (e_{H}) = e_{G} ϕ (e_{H})$ since $e_{G}$ is the identity, by right cancellation this yields $e_{G} = ϕ (e_{G})$ as needed.

Powers Pass Through a Homomorphism

Let

(G, ⋆_{G})

and

(H, ⋆_{H})

be two groups and

ϕ

a homomorphism between them. Let

n \in ℤ

and let

g \in G

be any group element, then we have

ϕ (g^{n}) = ϕ {(g)}^{n}

We first show it for all $n \in ℕ_{1}$ via induction. For $n = 1$ we can see that $ϕ (g^{1}) = ϕ {(g)}^{1}$ trivially holds.

Ho we let $k \in ℕ_{1}$ and assume that the statement holds for $k$ we'll now show it must hold for $k + 1$ , to do this we note the following : $\begin{matrix} ϕ (g^{k + 1}) & a m p; = ϕ (g^{k} ⋆_{G} g) \\ a m p; = ϕ (g^{k}) ⋆_{H} ϕ (g) \\ a m p; = ϕ {(g)}^{k} ⋆_{H} ϕ (g) \\ = ϕ {(g)}^{k + 1} \end{matrix}$ Where we've used our induction hypothesis going from the 2nd to the 3rd line.

The above holds us over for $ℕ_{1}$ , but it turns out we can now extend this to $ℤ^{-}$ quite easily, given some $z \in ℤ^{-}$ we first note that $e_{H} = ϕ (e_{G}) = ϕ (g^{z} ⋆_{G} g^{- z}) = ϕ (g^{z}) ⋆_{H} ϕ {(g)}^{- z}$ , note that we justify the last equality because $- z \in ℕ_{1}$ and then use our previous paragraph's proof. Finally by multiplying on the right by $ϕ {(g)}^{z}$ then we obtain $ϕ {(g)}^{z} = ϕ (g^{z})$ as needed.

We could stop there and we've left nothing out. Well indeed we have left nothing out, that is we've left $0$ out, well if $ϕ (g^{0}) = ϕ (e_{G}) = e_{H} = ϕ {(g)}^{0}$ as needed.

Powers pass through an Isomorphism

Let

(G, ⋆_{G})

and

(H, ⋆_{H})

be two groups and

ϕ

a isomorphism between them. Let

n \in ℤ

and let

g \in G

be any group element, then we have

ϕ (g^{n}) = ϕ {(g)}^{n}

We know that an isomorphism is a homomorphism, therefore powers pass through it