Consequence of First Homomorphism Theorem? Let $\phi:G\to\bar G$ be a surjective group homomorphism with kernel $N$. Then the first homomorphism theorem tells us that $G/N\cong\bar G$.
My question is this: Lagrange's theorem also tells us that $\frac{|G|}{|N|}=|\bar G|$ or $|G|=|\bar G||N|$ when $G$ is finite. Is it true that every element of $\bar G$ has exactly $|N|$ pre-images in $G$ under $\phi$?
For example, the identity of $\bar G$, call it $e_{\bar G}$, clearly has $|N|$ pre-images since $N$ is the kernel. We could say
$$\phi^{-1}(\{e_{\bar G}\})=N$$
Can always say that, if $\bar g\in \bar G$, then $|\phi^{-1}(\{\bar g\})|=|N|$
I'm still in my first semester of algebra, so keep it simple if possible. Thanks.
 A: The first isomorphism theorem indicates elements $\bar{g}\in\overline{G}$ have fibers $gN\in G/N$ (a fiber is the preimage of a single element). Note that if two different $g$s in $G$ represent the same $\bar{g}$ in $\overline{G}$, then they represent the same coset $gN$ in $G/N$ (and vice-versa), so this is sensible to say.
Every coset $gN$ of $N$ has the same size as $N$. Indeed for any coset $gN$ (with choice of representative $g$) there is an simple bijection $N\to gN$ given by $n\mapsto gn$.
A: The preimage of an element is a double coset of $N$. Specifically, 
$$\phi^{-1}(\phi(a))=NaN$$
So every element of the preimage is of the form $nan'$ for $n,n'\in N$. However, $N$ is normal, so since
$$nan'=a(a^{-1}na)n'$$
this double coset is also a left (and right) coset of $N$. So the preimage of an element $\phi(a)$ is $aN$ and hence has exactly $N$ elements.
(It is actually a bit more complicated than this; $\phi^{-1}(\phi(a))=\{n_1a_1n_2a_2\cdots n_ka_kn_{k+1}:n_1,\ldots,n_{k+1}\in N,\ a_1a_2\cdots a_k=a\}$, but by the same argument this is just the same left coset.)
