Is it true that $G\cong (G/H)\times H$ for every abelian group $G$ and every subgroup $H\le G\;?$

Question

Is it true that $G\cong (G/H)\times H$ for every abelian group $G$ and every subgroup $H\le G\;?$

So I know that by Lagrange's Theorem that $o(H)=o(G/H)= o(G)/o(H)$. I wanted to use a theorem that states that if two groups are cyclic (hence abelian), the order of two groups is equal if and only if they are isomorphic. So now I am stuck, I think that the two groups are isomorphic. Now I'm not sure where to go. I know that $o(H)$ divides $o(G)$. Would it suffice to show that $\big(o(G)/o(H)\big)\times o(H)= o(G)\;?$

Answer 1

Hint: there are two different abelian groups of order $4$, but only one of them can be $\Bbb Z_2\times \Bbb Z_2$.

Let $G$ be the other group of order $4$.

$G$ must contain a copy of $\Bbb Z_2$, and then you have $G/\Bbb Z_2 =\Bbb Z_2$ but $G\neq \Bbb Z_2×\Bbb Z_2$.

What is $G$?

Answer 2

Just because two groups have the same order does not make them isomorphic! Note that $\mathbb{Z}/4\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z},$ despite the fact that if $G = \mathbb{Z}/4\mathbb{Z}$ and $H = \{0, 2\}$ then $G/H \cong H \cong \mathbb{Z}/2\mathbb{Z}.$