Group Theory and Cosets. Want to show: 
Let G be a finite group and $N\leq G$. $\forall b\in N$, $bN=N$.
Proof:
Let $a \in bN$ that means $a=bn$, for some $n \in N$. Since $b \in N$ and since N is a group, it is closed and therefore $a=bn \in N$. This means that $bN \subseteq N$. Furthermore, since $|N|=|bN|$ (I proved this before attempting this exercise) it follows that $N=bN$.
Is the proof correct? I'm trying to improve my proof writing skills, and would very much like your feedback. Furthermore, i'd also like to see how you would write this proof, please.
My Questions:
1)For two sets, A,B, such that their cardinality is infinite (but the same),if $B \subset A$ does it follow that $A=B$ ? How would you prove this?
2) Can I just say that if b$\in N$ is arbitrary then $\{$ $bn: n\in N$ $\}$ $=$ $N$? I think I can because N is closed, and no matter what b I pick, multiplying it by every n in N  will give me the group? This is because G,N are finite and this is a direct consequence of the fact that if G is a group then every row and every column has distinct elements. Can this be generalized to infinite groups?
 A: $\newcommand{\N}{\mathbb{N}}$$\newcommand{\Z}{\mathbb{Z}}$I know your problem is stated for finite groups, but I think that this is misleading, because the result holds true in general, and the proof for the general case, as you will see in a minute, is not more difficult than the one for the finite case. (The cardinality argument fails when $N$ is infinite, as $2 \N \subsetneq \N$ provides an example of two sets with the same cardinality, one properly contained in the other.)
Once you have proved (without appealing to finiteness) that for $b \in N$ you have $b N \subseteq N$, just note that since $N$ is a subgroup, if $b \in N$ then $b^{-1} \in N$ as well, so that $b^{-1} N \subseteq N$ and thus, multiplying on the left by $b$, you obtain $N \subseteq b N$.
A: Its fine. Any way the other direction is easyly follows from group theoretic argument. Let $n \in N$. Already $b$ is in $N$, so $b^{-1}n \in N$. Thus $n=b(b^{-1}n) \in bN$, concluding $N \subseteq bN$.
For your first question, consider $B=\Bbb N$ and $A=\Bbb R$. Both are infinite and one is contained in the other. Does it means both are equal?

Can I say that if $b \in N$ is arbitrary then $bN=N$ ?

Why doubt about this? Its clear from the theorem in the OP
