If every element of $H$ and $G/H$ is a square, then prove that so is every element of $G$.

Question

Let $H$ be a subgroup of an abelian group $G$ such that every element of $H$ can be written as $b^2,\, b \in G$ and similarly for $G/H$. Then how to prove that every element of $G$ can also be written as a square ?

Answer 1

For each $g\in G$, There exists $a\in G$ such that $(aH)^2=gH$, which means that there exists $h\in H$ such that $g=a^2h$. Also, since there exists $h'\in H$ such that $(h')^2=h$, thus $$g=(ah')^2,$$ and the proof is completed.

Answer 2

HINT: Given any $a\in G$, you can write $aH = x^2H$ for some $x\in G$. Now proceed.

Answer 3

Hint: Note that $gH=aHaH=a^2H$ so elements of the coset $gH$ have the form...what?

Answer 4

Alternative proof. Apply the Snake Lemma to $$\begin{array}{c} 0 & \rightarrow & H & \rightarrow & G & \rightarrow & G/H & \rightarrow & 0 \\ & & 2 \downarrow ~~& & 2 \downarrow ~~ & & 2 \downarrow ~~ \\0 & \rightarrow & H & \rightarrow & G & \rightarrow & G/H & \rightarrow & 0 \end{array}$$