Index of a subgroup is preserved under group isomorphism

Question

Let $\varphi : G\to G'$ be a group isomorphism and let $N\lt G$.

prove: if $[G:N]\lt \infty$ then $[G:N]=[G':\varphi(N)]$

It is known that $\varphi(N)$ is indeed a subgroup of $G'$,

but how can I show equality of indices?

Note: I have yet to learn any isomorphism theorems.

Any help would be appreciated

Answer 1

Let $gN$ be a left coset for $g\in G$. Then, since $\varphi$ is a group homomorphism, if you take $n\in N$, then $\varphi(gn)=\varphi(g)\varphi(n)$, so clearly $\varphi(gN)\subseteq\varphi(g)\varphi(N)$. Since $\varphi$ is surjective, one has that every left coset of $G'$ is of the form $\varphi(g)\varphi(N)$.

Injectivity gives us that no two distinct cosets $g'N$ and $gN$ map to the same $\varphi(g)\varphi(N)$, since if $\varphi(g')\varphi(N)=\varphi(g)\varphi(N)$, then for some $n,n'\in N$

$\varphi(g')\varphi(n')=\varphi(g)\varphi(n).$

But this is the same as saying $g'n'=gn$, so $g'N=gN$.

Puting all together one concludes that the number of left cosets are the same.