Basic Theorems Regarding Group Homomorphisms

Recall from the Group Homomorphisms page that if $(G, \cdot)$ and $(H, *)$ are two groups then a homomorphism between these groups is a function $f : G \to H$ such that for all $x, y \in G$ we have that:

(1)

\begin{align} \quad f(x \cdot y) = f(x) * f(y) \end{align}

We will now look at some basic results regarding group homomorphisms.

Proposition 1: Let $(G, \cdot)$ and $(H, *)$ be two groups and let $f : G \to H$ be a homomorphism between these groups. Then: a) If $e$ is the identity in $G$ then $f(e)$ is the identity in $H$. b) For all $x \in G$, $f(x^{-1}) = [f(x)]^{-1}$. c) For all $x \in G$ and all $n \in \mathbb{Z}$, $f(x^n) = [f(x)]^n$.

• Proof of a) Let $e_1$ be the identity in $G$ and let $e_2$ be the identity in $G$. Then:

(2)

\begin{align} \quad e_2 * f (e_1) = f(e_1 \cdot e_2) = f(e_1) * f(e_1) \end{align}

• By cancellation, we have that $e_2 = f(e_1)$. $\blacksquare$

• Proof of b) Let $x \in G$ and let $e_1$ be the identity in $G$, $e_2$ be the identity in $H$. Then by part (a):

(3)

\begin{align} \quad f(x^{-1}) * f(x) = f(x^{-1} \cdot x) = f(e_1) = e_2 \end{align}

• Since the inverse of an element in a group is unique, the equation above tells us that $f(x^{-1}) = [f(x)]^{-1}$. $\blacksquare$

Proposition 2: Let $(G, \cdot)$ and $(H, *)$ be groups and let $f : G \to H$ be a homomorphism between them. If $H \subseteq G$ is a subgroup of $G$ then $f(H) \subseteq H$ is a subgroup of $H$.

• Proof: From the Subgroups and Group Extensions page, to show that $(f(H), *)$ is a subgroup of $(H, *)$ we must show that $f(H)$ is closed under $*$ and contains its inverses.

• Clearly $f(h)$ is closed under $*$. If not then there would exist $f(x), f(y) \in f(H)$ such that $f(x) * f(y) \not \in f(H)$. But then $f(x*y) \not \in f(H)$ which is a contradiction since $x, y \in H$ and $x * y \in H$ since $H$ is a subgroup of $G$.

• Since $H$ is a subgroup of $G$, for each $x \in H$ there exists an $x^{-1} \in H$ such that $x \cdot x^{-1} = e_1$ and $x^{-1} \cdot x = e_1$. From Proposition 1 above we see that for all $f(x) \in f(H)$:

(4)

\begin{align} \quad f(x) * [f(x)]^{-1} = f(x) * f(x^{-1}) = f(x \cdot x^{-1}) = f(e_1) = e_2 \end{align}

(5)

\begin{align} \quad [f(x)]^{-1} * f(x) = f(x^{-1}) * f(x) = f(x^{-1} \cdot x) = f(e_1) = e_2 \end{align}

• Therefore $f(H)$ is a subgroup of $H$. $\blacksquare$

Proposition 3: Let $(G, \cdot)$ and $(H, *)$ be groups and let $f : G \to H$ be a homomorphism between them. If $H \subseteq H$ is a subgroup of $H$ and if $f^{-1}(H) \neq \emptyset$ then $f^{-1}(H) \subseteq G$ is a subgroup of $G$.

The notation "$f^{-1}(H)$" denotes the inverse image of the set $H$ in $H$. It is given as $f^{-1} (H) = \{ x \in G : f(x) \in H \}$.

• Proof: By the other proposition on the page linked above, we need to show that for all $x, y \in f^{-1}(H)$ that $x \cdot y^{-1} \in f^{-1}(H)$ to show that $f^{-1}(H)$ is a subgroup of $G$.

• Let $x, y \in f^{-1}(H)$. Then $f(x), f(y) \in H$. Since $H$ is a subgroup of $G_{2}$, $[f(y)]^{-1} \in H$. But by Proposition 1, $[f(y)]^{-1} = f(y^{-1}) \in H$ and since $H$ is closed under under $*$ we have that $f(x) * f(y^{-1}) = f(x \cdot y^{-1}) \in H$. So $x \cdot y^{-1} \in f^{-1} (H)$ and $f^{-1} (H)$ is a subgroup of $G$. $\blacksquare$

Corollary 4: Let $(G, \cdot)$ and $(H, *)$ be groups and let $f : G \to H$ be a homomorphism between them. If $H \subseteq H$ is a NORMAL subgroup of $H$ and if $f^{-1}(H) \neq \emptyset$ then $f^{-1}(H) \subseteq G$ is a normal subgroup of $G$.