Please check proof of Lagrange's theorem I would appreciate it if someone could check this proof of Lagrange's theorem using cosets:
$G$ a group, $H$ a subgroup of $G$, and $C$ a left coset of $H$. We will show $|C|=|H|$. 
Take an $x\in C$, and consider $S_1=xH$. Then clearly $|S_1|=|H|$*. Take any element $y\in S_1$, say $y=xh_1$ for some $h_1\in H$. Then
$$yH=\{yh|h\in H\}=\{xh_1h|h\in H\}=^{**}xH$$
So $y\in C$, so $|C|\geq|S_1|=|H|$.
Now consider $S_2=C^{-1}x$, then clearly $|S_2|=|C|$. Also, by the definition of the coset, $S_2\subset H$, so $|C|=|S_2|\leq|H|$
Which proves $|C|=|H|$.

*It is clear that $|S_1|\leq |H|$. Now if $xh_1=xh_2$, then by the cancellation law $h_1=h_2$, so the $xh$ are all different, hence $|S_1|\geq |H|$.
**Take any $h\in H$. Then $h_1(h_1^{-1}h)=h$, and since $H$ is a subgroup $h_1^{-1}h\in H$. Also, again because $H$ is a subgroup, $h_1h\in H$ for all $h\in H$. This shows that $h_1H=H$ for $h_1\in H$

Based on comments, another version:
Take an $x\in C$, and consider $S_1=xH$. Then clearly $|S_1|=|H|$, and since $S_1=C$, $|C|=|S_1|=|H|$

This is a lot of good thinking. It is not a proof yet. What you've shown is that all the cosets are the size of $H$, but in order to deduce Lagrange's theorem, you need to show two other things: (1) the distinct cosets are disjoint from each other; (2) they cover $G$. Not unless you convince your audience of these two facts will you have shown that the group is partitioned into cosets of equal size.
(Aside: your proof that $|C|=|H|$ is correct, but actually quite a bit longer than it needs to be. Note that $C$, as a left coset, has the form $xH$ for some $x\in G$. Your argument that $|S_1|=|H|$ based on $S_1 = xH$ did not use anything special about $x$, and therefore the same argument applies without change to $C$ in the first place. You were basically done already once you could assert $|S_1|=|H|$! [Edit: you've noticed this in your edit!] On the other hand, to prove (1) mentioned above, you will need to argue along the same lines as some of your extras.)
