Question

Prove that every homomorphic image of a metabelian group is metabelian.

Short answer

It is proved that, every homomorphic image of a metabelian group is metabelian.

Step-by-step solution

Step by step solution Step 1: Important Theorems

First Isomorphism Theorem

Let $f:G\to H$ be a surjective homomorphism of a group with kernel K. Then, the quotient group $G/K$ is isomorphic to H.

Third Isomorphism Theorem

Let K and N be normal subgroups of a group G with$N\subseteq K\subseteq G$ . Then, $K/N$ is a normal subgroup of $G/N$, and the quotient group $(G/N)/(K/N)$ is isomorphic to $G/K$.

Proving that every homomorphic image of a metabelian group is metabelian

Suppose G is a metabelian group with subgroup N such that N is abelian.

Therefore, it is normal in G and $G/N$ is abelian.

From First Isomorphism Theorem, the homomorphic image of G is isomorphic to the quotient group $G/K$ for some normal subgroup K of G.

Since N and K are normal subgroups of group G, therefore for any arbitrary $n\in N,k\in K$ and $\mathrm{a}\in \mathrm{G}$, $a\left(nk\right)a^{-1}=an{a}^{-1}ak{a}^{-1}\in NK$.

So, NK is also normal.

Therefore, from theThird Isomorphism Theorem,$(G/K)/(NK/K)\cong G/NK$

This implies $(G/K)/(NK/K)$ is abelian as it is isomorphic to the homomorphic image of role="math" localid="1652771718971" $G/N$and we assumed that $G/N$is abelian.

From Second Isomorphism Theorem, $NK/K\cong N/(N\cap K)$

Therefore, by First Isomorphism Theorem $NK/K\text{\hspace{0.17em}}$ is a homomorphic image of abelian group N, which implies that $NK/K\text{\hspace{0.17em}}$is also abelian.

Since NK is a normal subgroup of G, so $NK/N$ is normal to $G/K$.

Thus, we have shown that $G/K$ is a metabelian group that contains subgroup $NK/N$ and it is a homomorphic image of group G.

Hence, it is proved that every homomorphic image of a metabelian group is metabelian.