# Proof of coset/subgroup property

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?

• It’s not “the identity from $H$”, it’s the identity of the group. Recall that a subgroup of $G$ must contain the identity element of $G$. Now, which cosets of $H$ in $G$ contain the identity? $H$ does; but in fact, it’s the only coset that does, because distinct cosets are disjoint. So no other coset contains $e$, so no other coset can possibly be a subgroup of $G$. So the only coset that could possibly be a subgroup is $H$ itself, which in fact is a subgroup. – Arturo Magidin Apr 24 at 5:44
• Each element of the group $G$ is an element of exactly one left coset of $H$. A subgroup of $G$ must contain the identity element of $G$, so there is just one left coset of $H$ that could possibly be a subgroup, the left coset containing the identity of $G$. But that coset is $H$ which really is a subgroup. – Lord Shark the Unknown Apr 24 at 5:53

They're trying to find cosets in which the identity from $$H$$ will not be present.

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).

The second line is based (sic)off of the first line.

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 $$G$$".

"It seems line they are magically jumping to it".

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.