# Trying to prove an isomorphism theroem

I am going through the following book on abstract algebra by Paul Garret .

http://www-users.math.umn.edu/~garrett/m/algebra/notes/Whole.pdf

At page 38 , the following theorem is illustrated :

"Given a normal subgroup , $$\ N$$ , of a group $$\ G$$ . and another subgroup $$\ H$$ ,of it ,and $$\ q$$ be the map $$\ q : G -> G/N$$ , then

$$\ H.N$$ = $$\{ hn : h \in$$ $$\ H$$ , $$g \in$$ $$\ G$$ } = $$\ q\ ^{-1}$$($$\ q$$ (H))

is a subgroup of $$\ G$$ and if $$\ G$$ is finite then the order of the group is

$$\ |H . N|$$ = $$\ |H|$$ $$\ |N|$$ / $$\ |H$$ $$\ \cap$$ $$\ N$$ | .

$$\ q(H)$$ $$\ \cong$$ $$\ H$$ / $$\ H \cap$$ $$\ N$$ . "

So , I could prove the first part , for the second part about the cardinalities , I could not follow the proof in the book as such , I tired proving it myself . So I noticed that for the group $$\ HN$$ , can be partitioned as the elements in $$\ H$$ $$\ \cap$$ $$\ N$$ and the elements not in it .

As the elements in $$H \cap N$$ form a group , I can consider the $$H \cap N$$ as an coset with the identity element .

Now it is also a subgroup in $$\ H$$, sp the number of cosets of it in $$\ H$$ is $$\ |H| /|H \cap N$$ and similarly , $$\ |N| /|H \cap N$$ in $$\ N$$ .

Now , I intended to show that the cosets of the group $$\ H \cap N$$ in the group $$\ HN$$ can be generated by the group operations of the cosets of the groups $$\ H \cap N$$ in the groups $$\ H$$ and $$\ N$$ .

So , let $$h_1$$ be an element belonging to one such coset of $$\ H$$ , and $$\ n_1$$ be an element belonging to one such coset of $$\ N$$ .

so all the elements in that coset of $$H$$ can be expressed as $$\ h_1. k$$ where $$\ k$$ is some element belonging to $$h \cap n$$ .

and similarly all the elements in that coset of $$N$$ can be expressed as $$\ n_1. l$$ where $$\ l$$ is some element belonging to $$h \cap n$$ .

Now any element in $$\ HN$$ which is not in $$\ H$$ and not in $$\ N$$ can be expressed as $$hn$$ for some $$\ h$$ in $$\ H$$ and for some $$\ n$$ in $$\ N$$ where $$h$$ and $$n$$ belong to some coset in $$\ H$$ and $$\ N$$ respectively . As such the group operations of the cosets of $$\ H \cap N$$ in groups $$\ H$$ and $$\ N$$ yield the cosets of the group $$\ H \cap N$$ in $$\ HN$$ . As such the total number of elements in $$\ HN$$ is equal to the total number of cosets in it times the cardinality of $$\ HN$$ ,i.e., $$\ |H|/|H \cap N |$$ * $$\ |N|/|H \cap N |$$ * $$\ |H \cap N |$$ = $$\ |H|/|H \cap N |$$ * $$\ |N|$$ .

The last part in my argument about covering up the cosets of the group $$\ HN$$ is unconvincing for me .

So the following are my questions regarding this :

1. Is the strategy of my proof correct ? (Would be a great help if the flaws could be listed out )
2. Can I refine the proof with rigor ?(How to extend this idea to provide a convincing proof ?)

******post -edit ************* An attempt to add some more conviction to the argument . For every coset , in both $$\ H$$ and $$\ N$$ let one element be chosen to represent the whole coset . Let's pick an element from $$|HN|$$ , $$\ X$$ , such that $$\ x$$ $$\notin$$ $$\ H$$ and $$\notin$$ $$\ N$$ . So , $$\ x$$ can be expressed as $$\ ab$$ where $$\ a$$ $$\ in$$ $$\ H$$ and $$\ b$$ in $$\ N$$ . Now let $$\ a$$ and $$\ b$$ be the representative of the coset they are in . Now all the elements in $$\ HN$$ which can be generated by the product of elements in the coset represented by $$a$$ and $$\ b$$ can be represented as $$\ a*b*p$$ where $$\ p$$ is some element belonging to $$\ H \cap N$$. Let's choose $$\ ab$$ to be the representative of the coset in $$\ HN$$ it is in . Now , we need to show that another pair $$\ cd$$ with $$\ c$$ $$\ \in$$ $$\ H$$ and $$\ d$$ $$\ \in$$ $$\ N$$ cannot yield the same element . If $$\ ab$$ = $$\ cd$$ then $$\ ab d^{-1}$$ = $$\ c$$ ,which is not possible , as for that to happen $$\ b d^{-1}$$ mus be in $$\ H$$ . As such the product of the number of representative elements of the cosets in both $$\ H$$ and $$\ N$$ provide the number of cosets of $$\ NH$$ . Hence the total number of cosets is $$\ |H|/|H \cap N | * |N|/|H \cap N |$$

Once you know that $$HN$$ is a subgroup of $$G$$, you can consider it as a group on its own, and likewise for $$H$$. Now the trick is to observe that $$N$$ is a normal subgroup of $$HN$$ and to consider the map $$f\colon H\to HN/N,\qquad f(x)=xN$$ This is well defined, no problem about that, and a group homomorphism.
What's the kernel? We have $$f(x)=N$$ (the identity element in $$HN/N$$) if and only if $$x\in N$$. Therefore the kernel is $$H\cap N$$.
Now the homomorphism theorem provides a unique injective group homomorphism $$g\colon H/(H\cap N)\to HN/N$$ such that $$g(x(H\cap N))=f(x)=xN$$. Since the map $$f$$ is surjective (prove it), we conclude that also $$g$$ is surjective, hence an isomorphism.
As a consequence $$|H/(H\cap N)|=|HN/N|$$. But this, in view of Lagrange's theorem, translates to $$\frac{|H|}{|H\cap N|}=\frac{|HN|}{|N|}$$ and therefore we obtain $$|HN|=\frac{|H|\,|N|}{|H\cap N|}$$ as required.