# Group Theory - Subgroup Proof

1. Jan 29, 2013

### CAF123

1. The problem statement, all variables and given/known data
Let G be a finite group,
a)Prove that if $g\,\in\,G,$ then $\langle g \rangle$ is a subgroup of $G$.
b)Prove that if $|G| > 1$ is not prime, then $G$ has a subgroup other than itself and the identity.

3. The attempt at a solution

a) This one I would just like someone to check and make sure I haven't missed anything out. It is my first proof for group theory so if it looks sloppy anywhere, please comment.
Anyway, I have:

Suppose $g\,\in\,G$. Want to show that $\langle g \rangle ≤ G$.
Let $H = \langle g \rangle$. If $H$ is a subgroup, it satisfies being non-empty, closure and inverse.

Since $\langle g \rangle$ is a group, by definition, it must contain the identity. So it is definitely non-empty.
If $g_1, g_2\,\in\,\langle g \rangle,\,g_1 \neq g_2$, then $g_1 = g^a, g_2 = g^b, a \neq b$ since we want two distinct elements. The condition $0 < a < o(g), 0 < b < o(g)$ must also be met since if we include 0, we end up with the trivial case. So, $g^a g^b = g^{a+b}$ which is in $\langle g \rangle$, since it is cyclic $\Rightarrow$ a+b is some a+b mod (o(g)). so we have closure.

Lastly, If $h\,\in\,\langle g \rangle$, then $h\in\,(e,g^2,...,g^{o(g)-1}),$ so h = gk for some 0 ≤ k < o(g). Since $\langle g \rangle$ is a group, for any h in <g>, we have a h-1. More precisely, if G is finite then g-1= gk for some positive integer k. Hence there exists an inverse element for h. $\square$

2)I am a bit stuck on this one, but I have written something down:
I thought intially that the following statement would disallow the statement if |G| was prime, but I am having second thoughts. Here it is anyway:
If |G| = p, where p is a prime then G is cylic. In particular, G = <g>, and so by 1), <g> is a subgroup. However, this is equal to G. So it is not different from G and e. Hence, not a different subgroup.
(Now I am thinking why can I not take a subgroup of <g> such that it is not G or e?)

The rest of my thoughts are along these lines, but I am unsure of how to proceed.
Many thanks.

2. Jan 29, 2013

### micromass

Both proofs are wrong. Here is some criticism:

It might be good give the definition of $\langle g \rangle$. There are various (equivalent) definitions, and I would like to know which one you use.

You wrote: IF H is a subgroup THEN blablabla. But this is a totally useless statement. You actually have to PROVE that H is a subgroup. It is actually the converse statement that is of interest to you: IF it is nonempty, has closure and inverse, THEN it is a subgroups.

You used that $\langle g\rangle$ is a group. But this is exactly what you need to prove!!! So you can't use that!

Why is $a\neq b$ and why do we want two distinct elements?

OK, but I don't see how this restriction is useful.

OK, but you only need to prove that $g^ag^b\in \langle g\rangle$. I don't get why the statement of the modulus and the order is relevant.

Again, you used that $\langle g\rangle$ is a group. But this is what you need to prove. You can't assume that

You have to prove that IF |G| is not prime, then blablabla. The first thing you do is assume that |G| IS prime. This is clearly incorrect. We don't care about the case where |G| is prime, we only care about the case where |G| is not prime.

Your first line should be: assume that |G| is not prime. Then ....

3. Jan 29, 2013

### CAF123

First of all thanks for the detailed feedback.
The definition I have is that <g> = {e,g,g2,....,go(g)-1}
Ok.

I am proving that <g> is a subgroup, so if it is a subgroup it must be a group. Am I not allowed to assume it is a group? If not, why not?

I thought that was part of the definition. We want h1,h2 to be two elements in the group, so that then when we do something to these two elements, they remain in the group. But I suppose if we take gg = g2, $\neq$ g, so I see your point.

Okay, I will remove this. I put it in just to let the proofreader know that a+b is not completely arbritary.

Same as above? Can't I assume it's a group to show that it's a subgroup?

Can you give me a hint to get me started? Surely we consider the case of |G| being prime somewhere so we know the condition is not satisifed for prime |G|.
Many thanks.

Last edited: Jan 29, 2013
4. Jan 29, 2013

### micromass

You can't assume anything. You have to prove first that it is a group. After you have proven that it's a group, then you can use it in your proofs.

The only thing you can assume is that <g> = {e,g,g2,....,go(g)-1}, there is nothing else you know about <g>.

You never need to consider the case that |G| is prime. Why not? Because you only need to prove something for |G| nonprime.

As a hint: take G a group such that |G| is not prime. Take an arbitrary g in G with $g\neq e$. What can you say about o(g)? What can you say about <g>?

5. Jan 29, 2013

### CAF123

Am I right to say that |G| = o(g) only if G is a cyclic group? Is there something general you can say for o(g) for some g in general G?. I have proved in class that |<g>| = o(g).

A question relating to last post: when you say I need to prove that <g> is a group, should I include this in the actual proof above if I have already proved it elsewhere? If I make this correction and change the implications of my sentences, would everything be ok for 1) then?

6. Jan 29, 2013

### micromass

Think of Lagrange's theorem.

Yes, if you proved that it's a group, then it's ok.

7. Jan 29, 2013

### CAF123

By Lagrange's theorem, the order of <g> divides the order of G. So this implies |G| = k|<g>| = k(o(g)). The RHS is such that we don't have a prime. Do I now have to consider the prime decompositions of these two numbers?

8. Jan 29, 2013

### micromass

Ok, so we know that |<g>| divides |G|. There are two possibilities: either |<g>|=|G| or |<g>|<|G|. Right? You'll need to treat these two possibilities differently.

9. Jan 29, 2013

### CAF123

If |<g>| = |G| then |G| = o(g). So we can create subgroups with elements that are not necessarily the identity and G itself. If |<g>| < |G|, then G contains greater than o(g) elements. So G is not cyclic which implies |G| is not a prime. (by some theorem I have). This may be a step in the right direction, but I feel there is more needed.

Last edited: Jan 29, 2013
10. Jan 29, 2013

### micromass

Why?

So you proved that |G| is not prime? But this is the hypothesis we made from the start!! This is not what we want to prove!

Is this the first time you do proofs?? You might want to look through a proof book like Velleman.

11. Jan 30, 2013

### CAF123

If I take, for example something of the form {g,g-1}, am I right to say this is not a subgroup because if take the two elements and put them together, I don't get the an element in the group. If so, I see how my statement is false.

I am now considering saying something along the lines that if |<g>| = |G|, then there exists a bijection between these two sets. For the other case, |<g>| < |G|, we can't have surjectivity, but we can have injectivity. Would this help?

I have done proofs in induction/contradiction etc.. before but this type of proof, where I have to interconnect theorems in a precise order to reach some conclusion, I haven't had that much practice. I will take your suggestion of Velleman.

EDIT: By 1) provided that g is in G, then <g> is a subgroup of G. <g> is not the identity nor is it G itself unless |G| = |<g>|, in which case G is cyclic. If G is cylic, then it's order may/may not be prime(but for the purposes of this proof consider non-prime). (I am trying a slightly different approach - I think it is called the contrapositive). If |G| $\neq$ |<g>|, then |G| is not prime and so <g> qualifies as a subgroup. How about this? I worry again that I may have the inplication the wrong way round.

Last edited: Jan 30, 2013
12. Feb 7, 2013

### CAF123

I got this question back today. Apparently I wasn't allowed to assume <g> was a group( even though I am fairly sure we proved this in lecture).

Just to clarify then: To prove that <g> was a subgroup of G, I should first show that it obeys the 4 axioms (closure, associativity, inverse, identity) thereby showing it is indeed a group and then show it further satisfies the three axioms for a subgroup? All in the same proof?

Many thanks

13. Feb 7, 2013

### jbunniii

Associativity would be automatically inherited from G, so you would have to show that <g> contains the identity, that it is closed under the group operation, and contains inverses of all of its elements. This should be quite easy, assuming that <g> is defined to be the set of all $g^{n}$ where $n \in \mathbb{Z}$.

14. Feb 7, 2013

### CAF123

I am not entirely sure what you mean here. Don't I prove associativity to show that it is a group?

After this, having shown it is a group, I then use the test for a subgroup?

15. Feb 7, 2013

### jbunniii

If you take three elements, $a,b,c \in \langle g\rangle$, then these elements are also in $G$, and $G$ is a group, so associativity is true: $(ab)c = a(bc)$.
If you show that $\langle g\rangle$ is a group, and you know that $\langle g\rangle$ is a subset of the group $G$, then that means $\langle g\rangle$ is a subgroup of $G$.

16. Feb 7, 2013

### CAF123

. This is only true if G = <g>, is it?

This makes sense, but if I didn't know it was a subset, I would just use test for a subgroup after showing <g> was a group?

17. Feb 7, 2013

### jbunniii

No. It's true for any $a,b,c \in G$, so in particular it is true if $a,b,c \in \langle g \rangle \subseteq G$.
Well, if it is not a subset, then it cannot be a subgroup. After all, what is a subgroup except a subset which is a group?

18. Feb 8, 2013

### CAF123

Ok, so a subgroup is essentially a subset that is also a group. But if I didn't know it was a subset, should I prove this (which is extremely trivial, right? - since simply a cyclic group contains $g \in G)$and then say from here, this implies a subgroup.

Or alternatively perhaps, just prove it is a subgroup after I have proved it is a group?