Basic Theorems Regarding Group Homomorphisms

# Basic Theorems Regarding Group Homomorphisms

Recall from the Group Homomorphisms page that if $(G_1, *_1)$ and $(G_2, *_2)$ are two groups then a homomorphism between these groups is a function $f : G_1 \to G_2$ such that for all $x, y \in G_1$ we have that:

(1)\begin{align} \quad f(x *_1 y) = f(x) *_2 f(y) \end{align}

We noted that a homomorphism of a group is essentially an isomorphism without the restriction that $f$ is bijective. Thus a bijective homomorphism between groups is an isomorphism between those groups.

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

Theorem 1: Let $(G_1, *_1)$ and $(G_2, *_2)$ be two groups and let $f : G_1 \to G_2$ be a homomorphism between these groups. Then:a) If $e$ is the identity in $G_1$ then $f (e)$ is the identity in $G_2$.b) For all $x \in G_1$, $f(x^{-1}) = [f(x)]^{-1}$. |

**Proof of a)**Let $e_1$ be the identity in $G_1$ and let $e_2$ be the identity in $G_1$. Then:

\begin{align} \quad e_2 *_2 f (e_1) = f(e_1 *_1 e_2) = f(e_1) *_2 f(e_1) \end{align}

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

**Proof of b)**Let $x \in G_1$ and let $e_1$ be the identity in $G_1$, $e_2$ be the identity in $G_2$. Then by part (a):

\begin{align} \quad f(x^{-1}) *_2 f(x) = f(x^{-1} *_1 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 $f(x^{-1}) = [f(x)]^{-1}$. $\blacksquare$

Theorem 2: Let $(G_1, *_2)$ and $(G_2, *_2)$ be groups and let $f : G_1 \to G_2$ be a homomorphism between them. If $H \subseteq G_1$ is a subgroup of $G_1$ then $f(H) \subseteq G_2$ is a subgroup of $G_2$. |

**Proof:**From the Subgroups and Group Extensions page, to show that $(f(H), *_2)$ is a subgroup of $(G_2, *_2)$ we must show that $f(H)$ is closed under $*_2$ and contains its inverses.

- Clearly $f(h)$ is closed under $*_2$. 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 $G_1$.

- Since $H$ is a subgroup of $G_1$, for each $x \in H$ there exists an $x^{-1} \in H$ such that $x *_1 x^{-1} = e_1$ and $x^{-1} *_1 x = e_1$. From theorem 1 above we see that for all $f(x) \in f(H)$:

\begin{align} \quad f(x) *_2 [f(x)]^{-1} = f(x) *_2 f(x^{-1}) = f(x *_1 x^{-1}) = f(e_1) = e_2 \end{align}

(5)
\begin{align} \quad [f(x)]^{-1} *_2 f(x) = f(x^{-1}) *_2 f(x) = f(x^{-1} *_1 x) = f(e_1) = e_2 \end{align}

- Therefore $f(H)$ is a subgroup of $G_2$. $\blacksquare$

Theorem 3: Let $(G_1, *_2)$ and $(G_2, *_2)$ be groups and let $f : G_1 \to G_2$ be a homomorphism between them. If $H \subseteq G_2$ is a subgroup of $G_2$ and if $f^{-1}(H) \neq \emptyset$ then $f^{-1}(H) \subseteq G_1$ is a subgroup of $G_1$. |

*The notation "$f^{-1}(H)$" denotes the inverse image of the set $H$ in $G_2$. It is given as $f^{-1} (H) = \{ x \in G_1 : f(x) \in H \}$.*

**Proof:**By the other theorem on the page linked above, we need to show that for all $x, y \in f^{-1}(H)$ that $x *_1 y^{-1} \in f^{-1}(H)$ to show that $f^{-1}(H)$ is a subgroup of $G_1$.

- 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 Theorem 1, $[f(y)]^{-1} = f(y^{-1}) \in H$ and since $H$ is closed under under $*_2$ we have that $f(x) *_2 f(y^{-1}) = f(x *_1 y^{-1}) \in H$. So $x *_1 y^{-1} \in f^{-1} (H)$ and $f^{-1} (H)$ is a subgroup of $G_1$. $\blacksquare$

Corollary 4: Let $(G_1, *_2)$ and $(G_2, *_2)$ be groups and let $f : G_1 \to G_2$ be a homomorphism between them. If $H \subseteq G_2$ is a NORMAL subgroup of $G_2$ and if $f^{-1}(H) \neq \emptyset$ then $f^{-1}(H) \subseteq G_1$ is a normal subgroup of $G_1$. |