I am given a proof of the following property:
The only left coset of $H$ which is a subgroup of $G$ is $H$ itself.
The brief proof is based on the following properties:
1) $xH = H \iff x\in h$
2) $xH \cap yH =xH$ or $\varnothing$
I understand the proofs of the two above properties. Now the given proof of the coset/subgroup property goes as follows:
Since $e\in H$, $e\not \in gH$ for all $g \in G$ and $g\not \in H$ by properties 1 and 2.
$\therefore$ no coset of $H$ besides H contains $e$, the unique identity in $G$.
$\therefore H $ is the only coset of $H$ which is a subgroup of $G$.
I don't understand what is going on in the first line, it seems that they're trying to find cosets in which the identity from $H$ will not be present. The second line makes sense, it seems relatively obvious to me, though I assume it is based off of the first line… As for the third line it seems like they are 'magically' jumping to it, though I assume that may be because of my lack of understanding of the first line.
So, what is happening in this proof?
Best Answer
As a subgroup of $G$, "the identity of $H$" is the identity of $G$, namely $e$. Remember, every subgroup is a group in its own right, but also contains the identity of the larger group (as its own identity element).
Indeed, it is. I'll explain both lines, starting from the first.
Note that $e \in H$, so by property $1$ we have $eH = H$. Next, if $L$ is any other coset, then by property $2$, we have $L \cap H = \emptyset$. In other words, $e \notin L$. This is what the second line is saying in words : " no coset of $H$ besides $H$ contains $e$, the unique identity of $G$".
As I mentioned earlier, every subgroup of $G$ contains the identity element of $G$, so every coset other than $H$ doesn't have the identity so cannot be a subgroup!
This leaves $H$ as the only option, but we know that it is a subgroup, so the conclusion is that it is the only coset that is a subgroup as well.