Thread: Subgroups and Normal Subgroups

Well, my teacher has given us a study guide with no answers. I guess she is busy....anywho...if anyone could help I would really appreciate it.

Suppose that G is a group and that A is a normal subgroup of G.
1.) Suppose that A≤J≤G. Show that A is normal to J.
2.) Suppose that A≤J≤G. show that J/A ≤ G/A.

Attempt:

1.) Suppose that A≤J≤G (subgroup notation). We also suppose that A is a normal Subgroup of G. This means that for every x in G, xA = Ax. So really, all I have are definitions and assumptions and not sure where to go to from there. I was thinking along that lines that since A is normal to G it would automatically follow that A is normal to J since J is a subgroup of G and A is a subgroup of J.

2.) In order to prove that something is a subgroup of another then we have to show 3 things or two things. I also know that

J/A = { gA | g is in J } and G/A = {gA | g is in G}.....

a.) first we need to show that J/A is nonempty.
b.) x in J/A and y in J/A imply xy is also in J/A.
c.) x in J/A implies that x inverse is also in J/A.
If I could get any help I would really appreciate it.

Originally Posted by crystalwater

Well, my teacher has given us a study guide with no answers. I guess she is busy....anywho...if anyone could help I would really appreciate it.

Suppose that G is a group and that A is a normal subgroup of G.
1.) Suppose that A≤J≤G. Show that A is normal to J.
2.) Suppose that A≤J≤G. show that J/A ≤ G/A.

Attempt:

1.) Suppose that A≤J≤G (subgroup notation). We also suppose that A is a normal Subgroup of G. This means that for every x in G, xA = Ax. So really, all I have are definitions and assumptions and not sure where to go to from there. I was thinking along that lines that since A is normal to G it would automatically follow that A is normal to J since J is a subgroup of G and A is a subgroup of J.

You are on the right lines - basically, if you pick any element then . There is no reason you cannot just pick elements from .

Originally Posted by crystalwater

2.) In order to prove that something is a subgroup of another then we have to show 3 things or two things. I also know that

J/A = { gA | g is in J } and G/A = {gA | g is in G}.....

a.) first we need to show that J/A is nonempty.
b.) x in J/A and y in J/A imply xy is also in J/A.
c.) x in J/A implies that x inverse is also in J/A.
If I could get any help I would really appreciate it.

Well, you know that is non-empty and a group as from part (1), so these 3 things have already been proven.

What you need to show is that every element from is an element from .

Alrighty. The first one is a little clearer.

As for the second one, I was working on it and I said something along these lines. Also, if the first is proven correctly, then the second follows straight from that and all i have to do is prove that every element in J/A is an element in G/A?

Well, i said something like since g is in J and since also g is in G then it follows that g is both in J and G since they are just multiples of the set A. Do I need to think in different ways?

Again, thank you for the help so far.

Maybe you can do the problem for me mesmo and email it to me ....just kidding. I see both of your points. For the most part, I "think" i understand it, emphasis on think.

So, instead of picking elements in G, i would be picking elements in J? So if i prove that J is normal to G then A must be normal to J?

* You know, sometimes I think this kind of math just talks in circles sometimes. And because of that I'm never sure of what to do.*

Thanks guys.

Originally Posted by crystalwater

Maybe you can do the problem for me mesmo and email it to me ....just kidding. I see both of your points. For the most part, I "think" i understand it, emphasis on think.

So, instead of picking elements in G, i would be picking elements in J? So if i prove that J is normal to G then A must be normal to J?

* You know, sometimes I think this kind of math just talks in circles sometimes. And because of that I'm never sure of what to do.*

Thanks guys.

No, you know next to nothing about . If we take then clearly as . Thus, as this holds for all .