Subgroup Theorem: Proof and Explanation

[jeff1evesque]

Theorem:
A subset H of a group G is a subgroup of G if and only if:
1. H is closed under binary operation of G,
2. The identity element e of G is in H,
3. For all a $$\in$$ H it is true that $$a^{-1}$$ $$\in$$ H also.

Proof:
The fact that if H $$\leq$$ G then Conditions 1, 2, and 3 must hold follows at one from the definition of a subgroup.

Conversely, suppose H is a subset of a group G such that Conditions 1, 2, and 3 hold. By 2 we have the identity group axiom ($$e * x = x *e = x$$) fulfilled. Also, the inverse group axiom ($$a * a^{-1} = a^{-1} * a = e$$)is fulfilled by Condition 3 (from above). It remains to check the associative axiom. But surely for all a, b, c $$\in$$ H it is true that $$(ab)c = a(bc)$$ in H, for we may actually view this as an equation in G, where the associative law holds. Hence H $$\leq$$ G.

Question:
Can someone explain to me the statement above:

It remains to check the associative axiom. But surely for all a, b, c $$\in$$ H it is true that $$(ab)c = a(bc)$$ in H, for we may actually view this as an equation in G, where the associative law holds.

The statement in this proof assumes that the associativity is somewhat "obvious", can someone explain the nature of this concept?

Thanks,JL

 

[CompuChip]

Sure. If H is a group, then for any three elements a, b, c from H, it must be true that (ab)c = a(bc); that's the associative axiom.
But (ab)c in H is the same as (ab)c in G, because the operation in H is the same as that in G. Similarly, a(bc) is a product in H, but it is also a product in G.
Now in G, which is a group and therefore has an associativity axiom, it is true that (ab)c = a(bc). But again, since the multiplication is the same (i.e. the element you get by multiplying a with b, and then with c in G is the same element you get when you multiply a by c, then by c in H). So if (ab)c = a(bc) is true in G, it is true in H.

Perhaps think of an example. Consider the group of integers as subgroup of the real numbers. How do you show that the integers have an associative addition; for example show that 1 + (2 + 3) = (1 + 2) + 3? Well, if you view 1, 2 and 3 not as integers but as real numbers, then it is immediately obvious. Because for any three real numbers, a + (b + c) = (a + b) + c: they're a group.

 

[fmam3]

Just nitpicking here, but is writing something like $$H \leq G$$ valid? As an illustration, it's almost saying that $$(\mathbb{N}, +) \leq (\mathbb{R}, +)$$, which really does not make sense. Is it not clearer to write $$H \subseteq G$$?

 

[rasmhop]

Just nitpicking here, but is writing something like $$H \leq G$$ valid? As an illustration, it's almost saying that $$(\mathbb{N}, +) \leq (\mathbb{R}, +)$$, which really does not make sense. Is it not clearer to write $$H \subseteq G$$?

It does make perfect sense as long as we define $H \leq G$ to mean H is a subgroup of G. I believe Dummit-Foote does this so I guess it's fairly standard notation. I also recall seeing something similar in Hungerford's book, but perhaps it was simply $H \triangleleft G$ for normal subgroup. Note that $\leq$ is a partial order defined only on groups and that it requires H to be a group.

 

[fmam3]

It does make perfect sense as long as we define $H \leq G$ to mean H is a subgroup of G. I believe Dummit-Foote does this so I guess it's fairly standard notation. I also recall seeing something similar in Hungerford's book, but perhaps it was simply $H \triangleleft G$ for normal subgroup. Note that $\leq$ is a partial order defined only on groups and that it requires H to be a group.

Thanks for pointing that out --- my textbook (Gilbert) just uses the subset notation so I wasn't aware that you could write that. Thanks!

 

[jeff1evesque]

Sure. If H is a group, then for any three elements a, b, c from H, it must be true that (ab)c = a(bc); that's the associative axiom.

If you don't mind me asking, in the converse direction of the proof, where was it assumed that H is a group? I thought H was assumed to be a subset:

Conversely, suppose H is a subset of a group G such that Conditions 1, 2, and 3 hold.

A subset (in this scenario) is neither a subgroup or group- am I correct? Because if it was a group, then by definition, it's elements are associative, and I can see the conclusion to my question.

Thanks again,JL

 

[CompuChip]

Yes, that is the whole point. You assume that H is a subset (not (sub)group!) of G which satisfies the conditions 1, 2 and 3. And then you want to prove that H is in fact a subgroup (i.e. satisfies the axioms of a group).

By the way, $H \le G$ is standard notation for "H is a subgroup of G".

 

[jeff1evesque]

Yes, that is the whole point. You assume that H is a subset (not (sub)group!) of G which satisfies the conditions 1, 2 and 3. And then you want to prove that H is in fact a subgroup (i.e. satisfies the axioms of a group).

Oh cool, that's what I thought. However, back to my original question-- how do we prove the associative axiom? Since we cannot assume our subset H is subgroup (or group)- since that is what we are trying to prove- we cannot assume it has the characteristics of associativity. I am guessing somehow the three Conditions (maybe the closure) allows us to construct the reasoning for associativity?

Thanks,JL

 

[Hurkyl]

However, back to my original question-- how do we prove the associative axiom? Since we cannot assume our subset H is subgroup (or group)-

What does H have to do with whether or not the operation is associative?

 

[jeff1evesque]

What does H have to do with whether or not the operation is associative?

It's closed under the given binary operation?

 

[Hurkyl]

It's closed under the given binary operation?

Is that relevant to whether or not G's binary operation is associative?

 

[jeff1evesque]

What does H have to do with whether or not the operation is associative?

Good question, I think that's what I am trying to ask.

 

[CompuChip]

Oh cool, that's what I thought. However, back to my original question-- how do we prove the associative axiom? Since we cannot assume our subset H is subgroup (or group)- since that is what we are trying to prove- we cannot assume it has the characteristics of associativity.

No, but it follows, because H is a subset of G and G is a group.
Any three elements of H are also elements of G, and all elements of G -- in particular those that are also in H, multiply associatively.

 

[Hurkyl]

Good question, I think that's what I am trying to ask.

And the answer is that it doesn't: G's binary operation doesn't know or care about the set H.




I suppose it's worth mentioning all the detail if you are paying careful attention to types. If (G,*) is a group, and H is a subset of G, then we're defining a new binary partial operation @ on H by the definition

h @ h' := h * h'​

and wondering if (H,@) is a subgroup of G.

Now, if we plug this definition into the associativity axiom...


I should point out that this level of precision is typically more pedantic than useful... but it can be a lifesaver when things get confusing.

 

[jeff1evesque]

No, but it follows, because H is a subset of G and G is a group.
Any three elements of H are also elements of G, and all elements of G -- in particular those that are also in H, multiply associatively.

Got it, it is associative since H is a subset of G, and elements in G are associative. This is explicitly shown by Hurkyl- which shows elements of H are associative by the operation from G. Along with the two other (already proven group) axioms, H is clearly a subgroup of G.

If (G,*) is a group, and H is a subset of G, then we're defining a new binary partial operation @ on H by the definition

h @ h' := h * h'​

and wondering if (H,@) is a subgroup of G.

Now, if we plug this definition into the associativity axiom...


I should point out that this level of precision is typically more pedantic than useful... but it can be a lifesaver when things get confusing.

Thanks so much.

 

[CompuChip]

Yes. In a sense you can say that H "inherits" the associativity from G.
Maybe it's also interesting to point out, that this is the case for any subset H of G. So as long as H is a subset of G, for any elements a, b, c from H it is true that a(bc) = (ab)c, simply because that is true in G.
For example, if you consider in the real numbers under addition the subset of all odd numbers, then it is still true for example that (3 + 5) + 1 = 3 + (5 + 1).

Usually the "problem" is in the "closedness" of the operation. For example, the odd numbers are associative because the real number are, but they are not a group because 3 + 1 is not odd (then again, the even numbers are, as is any group that consists of (fixed) multiples of integers).