Let $G$ be a group, and $N,N'$ subgroups. I'm working on a proof, where I want to show that $N\triangleleft G$. In this proof I have shown that $N'\triangleleft G$, $N'\subset N$ and $[G:N']=[G:N]=2$. Intuitively, I thought I could conclude that $N'=N$. However, I am in doubt now. This implication is trivial for finite $G$ (follows directly from Lagrange's theorem), but I don't know how I could proof this for the case $G$ is infinite. This made me doubt whether this is even true.
Using the identity $[G:N']=[G:N]\cdot[N:N']$ also does not seem to work since it assumes $G$ to be finite (right?). EDIT 2 Following the comments, this seems to be true for infinite $G$ also, so this proves the claim.
EDIT It might be important to note I need this to proof that a subgroup of index 2 is normal, so I cannot use this fact!
Is this true? Is it trivial and am I missing something? How would I proof this claim?
I tried searching this site already but I couldn't find anything related.
Since $[G:N]=2$, $N$ has two cosets in $G$. In particular, fix $g\in G\setminus N$, then the two cosets are $N$ and $gN$. Since $[G:N']=2$, $N'$ also has two cosets in $G$. Moreover, since $N'\subset N$, $(G\setminus N)\subset (G\setminus N')$. Therefore, $g\in G\setminus N'$ and the two cosets of $N'$ are $N'$ and $gN'$.
Since $N'\subset N$, $gN'\subset gN$. However, since the cosets form a partition, $G\setminus N=gN$ and $G\setminus N'=gN'$. Therefore, we have $gN\subset gN'$ from the argument above. Therefore, $gN'=gN$.
We observe that the multiplication by $g$ map $\mu_g:G\rightarrow G$, $\mu_g(g')=gg'$ is a bijection. Since, $gN=gN'$, so $N=N'$.