Consequence of First Homomorphism Theorem?

Question

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.

Answer 1

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 $n\mapsto gn$.

Answer 2

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.)