Show that if $G_{1}$ is a subgroup of $G$, there may be no subgroup of $G_{2}$ of $G$ such that $G=G_{1}\oplus G_{2}$.

Question

This question is Munkres' book Section 67 Exercise 2:

Show that if $G_{1}$ is a subgroup of $G$, there may be no subgroup of $G_{2}$ of $G$ such that $G=G_{1}\oplus G_{2}$. (Hint: Set $G=\mathbb{Z}$ and $G_{1}=2\mathbb{Z}$)

I have some attempts but I don't know how to proceed:

We set $G:=\mathbb{Z}$ and $G_{1}:=2\mathbb{Z}$ as suggested. Suppose there is a subgroup $G_{2}$ of $G$ such that $G=G_{1}\oplus G_{2}$. Then we know that $G/G_{2}\cong G_{1}$, which implies that $G_{2}$ contains some odd number $z$, since $G_{1}=2\mathbb{Z}$.

What should I do next to derive a contradiction?

Thank you!

Answer 1

Suppose $\Bbb Z=2\Bbb Z\oplus H$ for some subgroup $H$ of $\Bbb Z$. Then, for every $n\in \Bbb Z$ we have unique $k\in \Bbb Z$ and unique $h\in H$ such that, $n=2\cdot k+h$. Hence, for any element $h\in H$ we have, $2\cdot h+0=2h=2\cdot 0+2h$. Using uniqueness we have, $h=0$. That's $H$ is trivial group. $H$ is trivial means $\Bbb Z=2\Bbb Z\oplus\{0\}$, that is each element of $\Bbb Z$ is in $2\Bbb Z$, which is impossible as not all element of $\Bbb Z$ is even.

Answer 2

Perhaps an easier example is $G = \mathbb{Z} / 4 \mathbb{Z} = \{0,1,2,3\}$, with subgroup $H = 2\mathbb{Z} / 4\mathbb{Z} = \{0,2\}$.

But none of $\{0,1\}$,$\{0,3\}$ or $\{0,1,3\}$ form a subgroup, so there is no subgroup which intersects $H$ trivially, so no subgroup which is a complementary direct summand to $H$ in $G$.