Question

An element b of a group is said to be a square if there is an element c in the group such that$\mathit{b}\mathbf{=}{\mathit{c}}^{\mathbf{2}}$. Let$\mathit{N}$ be a subgroup of an abelian group G. If both$\mathit{N}$ and $\mathit{G}\mathbf{/}\mathit{N}$have the property that every element is square, prove that every element of $\mathit{G}$ is a square.

Short answer

It is proved that everyelement of$G$ is a square.

Step-by-step solution

As given in the question.

N is a subgroup of an abelian group G

N and G/N have the property that every element is square.

Proving that every element in G is a square.

Assume a to be an arbitrary element in G,$a\in G$.

We know that $G/N$has the property that every element is square.

Therefore, for any $b\in G$

$aN=N{b}^2\Rightarrow b^{-2}a\in N$

Again, we know that N has the property that its elements are square

So, for any $n\in N$

$\begin{array}{l}{b}^{-2}a=n^2\\ a=b^2{n}^2\end{array}$

Because$a\in G$ and G is abelian, so$a=b^2{n}^2={\left(bn\right)}^2$

Since a is an arbitrary element of G. Thus, from above it can be seen that every element of G is a square.