# Why HH=H?

1. Oct 20, 2006

### pivoxa15

Let G be a group and H a subgroup of G.

The book claims HH=H because H is a subgroup.

Group multiplication is defined as AB={(a,b): a in A, b in B}

So HH should be ordered pairs with each pair containing two identical elements in H. But why is the answer H, which is not an ordered pair?

I think they have used this definition http://en.wikipedia.org/wiki/Product_of_subgroups instead of http://en.wikipedia.org/wiki/Direct_product_(group_theory)

Are the two completely different? The latter they direct product. If one write HH does it not refer to direct product? I always thought not putting a sign such as X means the same thing as putting X. Or is this convention only for elements of a group. So when doing operations on whole groups, putting or not putting a sign has different consequences?

Last edited: Oct 20, 2006
2. Oct 21, 2006

### murshid_islam

$$HH = \left\{ h_{1}h_{2}\ |\ h_{1}, h_{2} \in H \right\}$$
$$h_{1}h_{2} \in H$$ since H is a subgroup and hence closed
therefore, $$HH \subseteq H$$

now let $$h \in H$$
$$h = he \in HH$$ where e is the identity element
since $$h \in HH, H \subseteq HH$$

we have $$HH \subseteq H$$ and $$H \subseteq HH$$
therefore,
$$HH = H$$

Last edited: Oct 21, 2006
3. Oct 21, 2006

### pivoxa15

I understand that. To show equality between two relations, do I always have to prove if and only if?

I take it that direct product is entirely different to product of subgroups? Couldn't you have a direct product between subgroups?

4. Oct 21, 2006

### murshid_islam

that is the definition of cartesian product, isn't it? here HH is not a cartesian product

5. Oct 21, 2006

### matt grime

They are groups, so of course you can have direct products. However, if H and K are subgroups of G, then in general HxK will not be a subgroup of G.

6. Oct 21, 2006

### pivoxa15

So as a general rule, always use X when meaning direct product between groups and no symbol when meaning subgroup multiplication which is not in Cartesian coordinate space. Correct?

Also with the proof when assuming product of subgroups what about

H is a subgroup of G.
$$H=\left\{h_{1}, h_{2}, ..., h_{n}\}$$

$$HH = \left\{ h_{1}h_{1}, h_{2}h_{2}, ..., h_{n}h_{n}\}$$

use close under multiplication for products of elements in subgroups.

$$HH = \left\{ h_{1}, h_{2}, ..., h_{n}\} = H$$

Last edited: Oct 21, 2006
7. Oct 21, 2006

### matt grime

Closure only tells you that HH is a subset of H. From what you have written, you apparently are saying that the elements of HH are of the for hh for h in H. You're set sould be, if you want to needlessly introduce subscripts:

$$\{ h_i h_j : 1 \le i,j \le n\}$$

not, as you have written

$$\{ h_i h_i : 1\le i \le n\}$$

I hope you see that these sets will probably be different. In fact the second set will only be H if the map sending x to x^2 is a bijection on H.

In general, if H and K are subgroups of G then HK is *not* a subgroup as well, by the way.

Last edited: Oct 21, 2006
8. Oct 21, 2006

### pivoxa15

murshid_islam proved that sending x to x^2 is a surjection on H didn't he but not necessarily an injection?

9. Oct 22, 2006

### matt grime

No. It is neither surjective, nor injective, in general.