# Question about Cosets and Lagrange's Theorem

1. Dec 22, 2014

### PsychonautQQ

If a is an element of G and H is a subgroup of H, let Ha be the right coset of H generated by a.

is Ha a subgroup?

I have this question because i feel like the answer should be know, yet my textbook notation makes it look like it is.

Why I think it should only be a set:
Let G = <a> and H = <a^3>. and |a| = 6,.
Then H = {1,a^3}, Ha = {a,a^4} Ha^2 = {a^2,a^5}.
To me it looks like Ha and Ha^2 can not be subgroups because they are not closed under the operation, unless the operation is multiplying the elements by a^3.. I don't know i'm confused.

But then I think they should be considered subgroups, because it goes on to say that Ha = H iff a is in H. So Ha must be a subgroup if it equals a subgroup... Anyway, anyone have insight here?

2. Dec 22, 2014

### Stephen Tashi

The problem doesn't not say that $a$ must be an element of $H$. What the textbook goes on to say about the case when $a$ is an element of $H$ is not relevant to the problem.

3. Dec 23, 2014

### jbunniii

$Ha$ is a subgroup if and only if $Ha = H$ if and only if $a \in H$.

One easy way to see this is that if $Ha \neq H$ then $Ha$ and $H$ are disjoint, so $Ha$ does not even contain the identity element.

4. Dec 24, 2014

### PsychonautQQ

so when you create G out of disjoint cosets, you are joining H (a subgroup of G) with Ha, Hg, Hh, etc etc, which are only disjoint sets of elements of G?

5. Dec 24, 2014

### jbunniii

Yes, $G$ is the disjoint union of $H$ and the other cosets of $H$. Only $H$ is a subgroup. The other cosets are subsets of $G$ with the same size as $H$, but they aren't subgroups.