Index and normal subgroups

Question

I want to show the following.

For an infinite group G with only two normal subgroups (G and {e}) holds: There does not exist a non-trivial subgroup of G with finite index.

I think i should prove this by contradiction. So there exists a partition of G:

Let U be a non-trivial subgroup of G, it holds: G=$g_1U\cup g_2U\cup...\cup g_nU$ for some $g_1,...g_n$. Maybe there is a possibility to show, that the gU are finite?

Answer 1

Hint: Suppose $G$ has a subgroup $H$ of finite index, then let $G$ act on the left cosets of $H$ by left translation. Look at the kernel of this action.

Answer 2

I just want to complete this prove for others and for completeness: As stated from me above there are only two possibilities for $Core_G(U)$ First: $Core_G(U)=\{e\}$ and second: $Core_G(U)=G$, because G has only two normal subgroups (G and $\{e\}$) and $Core_G(U)$ is clearly a normal subgroup of G. We want now get a contradiction:

1) If $Core_G(U)=\{e\}$ We get a contradiction because then G is finite by lagrange's theorem.

2) If $Core_G(U)=G$ then we want to show that $U=G$ which would be a contradiction to the chose of U and we where finish (We chose U as non-trivial)

Prove by contradiction: If $U\neq G$ then there is an element $h$ with $h \in G$ and $h\notin U$. By definition of the Core it holds $h\in gUg^{-1}$ $\forall g \in G$, so especially for $g=e$. But then $h\in U$ with is a contradiction.