Homomorphisms and Subgroups

Homomorphisms

Let $(G, \cdot), (H, *)$ be groups (where $\cdot, *$ are the group operations, respectively). We say $f: G \rightarrow H$ is a homomorphism if for all $x, y \in G$ $f(x\cdot y) = f(x)*f(y)$. Proposition: let $f: G \rightarrow H$ be a homomorphism. 1) $f(1_G) = 1_H$ (when we apply f to the identity element of G we get the identity element of H)

2) $\forall x \in G, f(x^{-1}) = f(x)^{-1}$ (here $\forall$ means for all)

Proof: 1) $f(1_G) = f(1_G \cdot 1_G) = f(1_G)*f(1_G)$ so $1_H = f(1_G)$ 2) $f(1_G) = f(x \cdot x^{-1}) = f(x)*f(x^{-1})$ so $f(x)^{-1} = f(x^{-1})$

An isomorphism is a bijective homomorphism.

Proposition: Let $f: G \rightarrow H$ be an isomorphism, then $f^{-1}:H \rightarrow G$ is an isomorphism.

Proof: $f^{-1}$ is a bijection (inverse of bijection is a bijection). Let $x, y \in H, x = f(f^{-1}(x)), y = f(f^{-1}(y))$ where $f^{-1}(x) = s \in G, f^{-1}(y) = t \in G$. Then $f^{-1}(x*y) = f^{-1}(f(s)*f(t))=f^{-1}(f(s\cdot t)) = s\cdot t = f^{-1}(x)\cdot f^{-1}(y)$

An automorphism is a group isomorphism from some group to itself. We denote the set of automorphisms of a group G by Aut(G).

Proposition: Aut(G) is a group under composition (here composition refers to function composition).

Proof: composition by definition is associative. The identity function $i:G\rightarrow G$ is a group automorphism so it is in Aut(G) and if $x, y \in G$ then $i(x\cdot y) = x\cdot y = i(x)\cdot i(y)$. If $f \in$ Aut(G), $f^{-1}$ is also an automorphism and $f^{-1}$ is the inverse of the group.

A monomorphism is an injective homomorphism. $f: G \rightarrow H$ is a group homomorphism, then f is a monomorphism if and only if $f^{-1}(1_H) = \{1_G\}$

Subgroups

Let G be a group and H a subset of G. We say H is a subgroup of G if H is nonempty and H is closed under products and inverses and we denote it $H \leq G$. In other words, $H \leq G$ if and only if $H \neq \{\varnothing\}$ and $\forall x, y \in H, x \cdot y^{-1} \in H$ (or $x \cdot y \in H$ if $H \leq G$ is finite).

Proposition: Let $f: G \rightarrow H$ be a homomorphism. Then $f(G) \leq H, f^{-1}(\{1+H\})\leq G$. More generally, if $K \leq G$ then $f(K) \leq H$ and if $L \leq H$ then $f^{-1}(L) \leq G$

Proof: $x, y \in f(K), x = f(s), y = f(t) \rightarrow s, t \in K$ $x\cdot y^{-1} = f(s)f(t)^{-1} = f(s\cdot t^{-1}) \in f(K)$. $x, y \in f^{-1}(L), f(x), f(y) \in L$. $f(x\cdot y^{-1}) = f(x)f(y^{-1}) \in L$.