Thread: 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 gh`is in H. Since h is in H, its inverse h`must 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 g`is 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.

"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."

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:

a+b (instead of a*b)
0 instead of e
-a instead of a^-1.

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).