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Problem 81 (The Image Of A Subgroup Under A Homomorphism Is A Subgroup)

Suppose that $f:G\to H$ is a homomorphism. Let $A$ be a subgroup of $G$. Show that $f(A)$ is a subgroup of $H$.

Solution

Let $G$, $H$ be groups. Let $f:G\rightarrow H$ be a homomorphism. Let $A\leq G$. We must prove that $f(A) \leq H$.

We will use the subgroup test. We must show $f(A)$ is nonempty, a subset of $H$, closed under the binary operation, and closed under taking inverses.

Since $f$ is a homomorphism we know $f(e_G) = e_H \in f(A)$. Thus $f(A)$ is nonempty.

Recall $f(A) = \{f(a) | a\in G\}$. By definition, $f(A)\subset H$ since $H$ is the codomain of $f$.

We claim that for every $c,d\in f(A)$ we have $c\cdot d \in f(A)$. Let $c,d\in f(A)$. This means $c=f(a)$ and $d=f(b)$ for some $a,b \in A$. We compute $$\begin{align} c\cdot d &= f(a)\cdot f(b)\\ &= f(a\cdot b), \end{align}$$ since $f$ is a homomorphism. Thus since $A\leq G$ we know $a\cdot b\in A$, which means $f(a\cdot b) \in f(A)$. This implies $c\cdot d \in f(A)$. Therefore $f$ is closed under the binary operation.

Finally, we claim that for every $c \in f(A)$ we have $c^{-1} \in f(A)$. Let $c\in f(A)$. Then $c=f(a)$ for some $a\in A$. We compute $$\begin{align} c^{-1} &= (f(a))^{-1}\\ &=f(a^{-1}). \end{align}$$ Since $A\leq G$, we know $a^{-1}\in A$. This implies $f(a^{-1})\in f(A)$, which means $c^{-1}\in f(A)$. Therefore $f$ is closed under taking inverses. Thus $f(A)$ is a subgroup of $H$. $\square$

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