# Thread: Having trouble understanding this theorem related to groups.

1. ## Having trouble understanding this theorem related to groups.

This is taken from Judson's Abstract Algebra textbook.

I will notate the inverse of h as h and the inverse of g as g.

"Let H be a subset of a group G. Then H is a subgroup of G if and only if H is non-empty and whenever g, h are elements of H then gh is in H."

Simple enough statement. The proof of it confuses me.

"Let H be a non-empty subset of G. Then H contains some element g. So gg=e is in H. If g is in H, then eg=g is also in H. Finally, let g, h be elements of H. We must show that their product is also in H. However, g(h)=gh is in H. Hence, H is indeed a subgroup of G. Conversely, if g and h are in H, we want to show ghis in H. Since h is in H, its inverse hmust also be in H. Because of the closure of the group operation, gh is in H."

This proof confuses me. It seems to just throw around random statements. I know what a subgroup is - it is subset of a group that is closed, contains the identity element, and is invertible. I don't understand how the above proof shows these things.

Another issue with this writing is that we are given "H is a non-empty subset that contains an element 'g.'" We are given no information other than "H is a subset." How do we conclude that gis also an element of this set?

Also, this statement bothers me:
"Finally, let g, h be elements of H. We must show that their product is also in H. However, g(h)=gh is in H."

I understand that we show that the product of g and h is in H to show that H is closed under the operation. I do not understand why we are working with "g(h)". Where did this "(h)" come from? The "however" also implies some kind of unexpected result.

Is this theorem trying to state the following?

gg is in H
hh is in H
g, g, h, h, e are in h

This implies that gh is in H.

The last statement is great, but nowhere did we state that H was a subgroup. gh must be an element of the group (G,+), but this proof that H is a subgroup seems to assume that H is a subgroup.

2. ## Re: Having trouble understanding this theorem related to groups.

"Let H be a non-empty subset of G. Then H contains some element. Let's call it g.

We are given that for any x, y ∈ H, xy⁻¹ ∈ H. (*)

First, we need to show that e ∈ H. Setting x = y = g in (*), we get
gg⁻¹ = e is in H.

Next, we want to show that for any g ∈ H, g⁻¹ ∈ H. Indeed, if g is in H, then setting x = e and y = g in (*), we get eg⁻¹ = g⁻¹ is also in H.

Finally, let g, h be elements of H. We must show that their product is also in H. However, setting x = g and y = h⁻¹ in (*), we get g(h⁻¹)⁻¹ = gh is in H. Hence, H is indeed a subgroup of G.

Conversely, if g and h are in H, we want to show gh⁻¹ is in H. Since h is in H and H is a subgroup, its inverse h⁻¹ must also be in H. Because of the closure of the group operation, gh⁻¹ is in H."

3. ## Re: Having trouble understanding this theorem related to groups.

In order to prove a subset H of a group G is in fact a group itself, we need to show that H satisfies the group axioms:

1. that * (the group operation of G) is a binary operation on H (this is where closure comes from).
2. that * is associative (but this is obvious, since * is associative on all of G, associativity clearly holds for any elements of G, even those in H). no need to prove this.
3. that H contains an identity. it's clear this must be the identity e of G. this also means H cannot be the empty set.
4. that every inverse of h in H is also in H (the inverse h-1 of h in H must be the SAME inverse as in G since it's the same operation *).

now, usually, we just show 1&4: because if H is closed under *, and has all inverses, then we have h*h-1 in H, so that we get 3 "for free".

however, the following condition is equivalent to 1 & 4 above:

1a. for ALL x,y in (a non-empty set) H, x*y-1 is in H.

obviously if 1 & 4 above hold, then y-1 is in H (by 4), so x*y-1 is in H (by 1).

so 1 & 4 --> 1a.

however, 1a --> 1&4 as well. the proof is, admittedly, a bit convoluted:

first we need to show 1a --> 4. well, to do THAT, we need to express for any h in H, h-1 in the form x*y-1. the logical choice is x = e, and y = h (which is what is typically done).

however, there's a catch: we haven't actually shown e is in H. so we need to do that first, before we can show 4 is derivable from 1a.

but, that's not hard to do: just pick some h in H (recall H MUST be non-empty), and set x = y = h, so that h*h-1 (= e, in G) is also in H.

NOW we can pick x = e, and y = h, which tells us that if h is in H, so is h-1 = eh-1.

closure is even a bit more convoluted to show, here we pick x = g, and y = h-1.

the reasoning behind this is the fact that (in G, and thus in H as well): (h-1)-1 = h. that is, if h-1 is the inverse for h, then h is the inverse for h-1: since h-1 is the UNIQUE element of G such that h*h-1 = h-1*h = e, and since (from the foregoing) we see that h is an element of G with: h-1*h = h*h-1 = e, so (by UNIQUENESS of inverses) h must be (h-1)-1.

now we already proved 1a--> 4, so from g,h in H, we have (from 4) that h-1 is in H, therefore:

g*(h-1)-1 = g*h is in H, which means that 1a implies 1 (closure) as well.

**********

the reason for doing all this, is it is often "easier" to show for x,y in H that x*y-1 is in H directly, instead of showing 1,3 & 4 separately. this is particularly so for ABELIAN groups, where instead of writing a*b, we typically write:

in this "additive notation", the "one-step subgroup test" becomes:

for a (non-empty) subset H of (G,+), H is a subgroup if and only if a - b is in H, for all a,b in H.

for example, we can show that nZ = {k in Z : k = nt, for some integer t} is a subgroup of (Z,+):

suppose a,b are in nZ. then a = nt, b = nu, for some integers t and u.

then a - b = nt - nu = n(t - u), which is in nZ (since t - u is an integer), so (nZ,+) is a subgroup of (Z,+).

note that we didn't need to go through the "extra steps" of showing -nt is in nZ, or that 0 (= n0) is in nZ (although to be fair, showing these two things isn't THAT hard).