# Isomorphic groups G and H, G has subgroup order n implies H has subgroup order n

• ArcanaNoir
In summary, G is isomorphic to H and if G has a subgroup of order n then H has a subgroup of order n. The Attempt at a Solution suggested that if g is an element in the subgroup of G then g maps to some h in the new subgroup of H. However, the proof is not rigorous and needs to start with the definition of closure.
ArcanaNoir

## Homework Statement

G is isomorphic to H.
Prove that if G has a subgroup of order n, H has a subgroup of order n.

## Homework Equations

G is isomorphic to H means there is an operation preserving bijection from G to H.

## The Attempt at a Solution

I don't know if this is the right approach, but
G has a subgroup of order n implies e is in the subgroup, and since e in G maps to e in H, e is in the potential new subgroup of H.
Suppose g is an element in the subgroup of G. Then, g maps to some h in the new subgroup of H.

Man, I think I'm proving the wrong thing here. I'm heading toward proving a subgroup in H, not that the order is the same. I need a new direction. Any one have a direction idea?

What is it?

Verify that the subgroup-conditions check out on the mapping of the subgroup.

A subgroup is a subset of a group which is closed on the group's operation, the identity in the group is in the subgroup, and each element in the subgroup's inverse is in the subgroup too.

Let the subgroup of G be g and the potential subgroup of H be h.
Since e is in any subgroup of G, $$\theta (e_g) = e_h$$ is in h.
$$\forall x \in g, \exists y \in h : \theta (x) = y$$ (because theta is bijection)
Since x is in g, x-1 is in g.
How do I show that this means y-1 is in h?
Also, how do I show a*b is in g implies c*d is in h?

Use $\theta(a*b)=\theta(a)*\theta(b)$.

Does $a*b \in g \Rightarrow \theta (a*b) \in h$ ?
Or perhaps $a, b \in g \Rightarrow \theta (a) \wedge \theta (b) \in h$?

Let's define h to be θ(g).

So if a*b is in g, then θ(a*b) is in h!

We know that g is closed, contains e, and contains the inverse of each of its elements.

Does h?

I like Serena said:
Let's define h to be θ(g).

Ah, the critical piece. Sometimes I lack a most basic but most critical piece. *computing*

θ(g) is certainly a better name for it than the "potential subgroup h"...

$$a,b \in g \Rightarrow a*b \in g \Rightarrow \theta (a*b) \in h \Rightarrow \theta (a)*\theta (b) \in h$$ is this enough to show h is closed? (oh, and pretend like I'm using the h operation for h.. I don't feel like going back and changing to a new symbol at this point)

$$a \in g \Rightarrow a^{-1} \in g \Rightarrow \theta (a^{-1}) \in h \Rightarrow (\theta (a))^{-1} \in h$$

Then, of course the number of elements is the same because theta is a bijection. So, All proved, right?

Hmm, it's sort of proved... but it's a bit sloppy and on an exam they like rigorous proofs...

To show h is closed, you need to start with the definition of closure, which is:
For each c and d in h, c*d is also in h.​

For each c and d in h, ...​
and you need to end with:
..., so c*d is in h.​

For each $a$ in h, ...​
and end with:
..., so there is a $b$ in h with $a*b = b*a = e$.​

That is, assuming you want a perfect grade!

Well, this is only studying so perfection isn't needed all the time. So long as I know what the idea is. I think I'm close enough on this one to move on comfortably.

One time I made the mistake of asking my prof about a homework problem, and showing him my homework pages (which we did not turn in). It was the tenth page of induction proofs, so I had stopped putting "assume this for k, that is assume (some equation) is true". He chastised me for my un-thoroughness. There just comes a point when ten pages into something I become un-thorough. :P

Plus, Latex takes so long to write. I think someday I'll get one of those fancy keyboards that you can program a bazillion functions into the buttons, and i'll make latex buttons.

I get it. You don't like to type.

I'd like to make another point though.
If you are wondering in a problem what it is exactly what you need to do, you usually need to drag up the exact definitions, and simply repeat them.
Then repeat them again, substituting different symbols into them as apply to the problem at hand.

From there, the solution often becomes suddenly obvious.

Btw, didn't you post in some thread this nifty tool that converts handwriting into latex? Or am I mixing this up?

Handwriting into latex? Nope, I didn't post anything like that.

The real problem with this section is our professor suddenly realized the semester is almost over and we have two tests to do yet, so he kind of rushed us on this section. He left out a lot of the types of problems I'm working on now. I'm an example-based learner.

Anyway, thanks for the help, as always :)

## 1. What is the definition of an isomorphic group?

An isomorphic group is a mathematical concept where two groups, G and H, have the same structure and are related through a bijective map that preserves the group operations.

## 2. How do you prove that two groups are isomorphic?

To prove that two groups, G and H, are isomorphic, you need to show that there exists a bijective map between them, called an isomorphism, that preserves the group operations. This means that for any elements a and b in G, the isomorphism must satisfy f(ab) = f(a)f(b). Additionally, you must also show that the inverse map also preserves the group operations, meaning that for any elements x and y in H, the inverse isomorphism must satisfy f^-1(xy) = f^-1(x)f^-1(y).

## 3. What does it mean for one group to have a subgroup of order n?

Having a subgroup of order n means that the group contains a subset of elements that form a subgroup with n elements. This subgroup must also satisfy the group axioms and have the same group operations as the larger group.

## 4. Does having a subgroup of order n in one group imply that the other group has a subgroup of the same order?

Yes, if two groups, G and H, are isomorphic and G has a subgroup of order n, then H must also have a subgroup of order n. This is because isomorphic groups have the same structure and thus must have the same subgroups.

## 5. How does the concept of isomorphic groups relate to the statement "G has subgroup order n implies H has subgroup order n"?

The statement "G has subgroup order n implies H has subgroup order n" is a consequence of two groups being isomorphic. If two groups are isomorphic, then any subgroups of one group must also have corresponding subgroups of the same order in the other group.

• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
793
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
2K