The Image Of A Subgroup Under A Homomorphism Is A Subgroup Mikai

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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

Suppose $f:G\rightarrow H$ is a homomorphism. Let $A$ be a subgroup of $G$.

We will show that $f(A)$ is a subgroup of $H$ using the subgroup test.
Recall $f(A)=\{f(A)|a\in A\}$. First, we will prove that $f(A)$ is nonempty. Since $A$ is a subgroup of $G$, then $e_G\in A$. This means $f(e_G)\in f(A)$.

We will now show $f(A)$ is closed under the binary operation of $H$.
Let $b,c\in f(A)$. We will show that $bc\in f(A)$. Pick $a_1,a_2\in A$ such that $f(a_1)=b$ and $f(a_2)=c$. From this we know that $bc=f(a_1)f(a_2)$. Since $f:G\rightarrow H$ is a homomorphism we know that $bc=f(a_1)f(a_2)=f(a_1a_2)$. This mean $bc\in f(A)$ since $a_1a_2\in A$.

We will now show that $f(a)$ is closed under taking inverses.
Suppose $b\in f(A)$. Pick $a\in A$ such that $b=f(a)$. Then this mean $b^{-1}=[f(a)]^{-1}=f(a^{-1})\in f(A)$ by problem 76 and since $a^{-1}\in A$.

Therefore, by the subgroup test $f(A)$ is a subgroup of $H$.

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• Week 11
• Mikai
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