# Thread: Subgroups and Normal Subgroups

1. ## 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.

2. 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 $\displaystyle x \in G$ then $\displaystyle x^{-1}Ax = A$. There is no reason you cannot just pick elements from $\displaystyle J$.

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 $\displaystyle J/A$ is non-empty and a group as $\displaystyle A \unlhd J$ from part (1), so these 3 things have already been proven.

What you need to show is that every element from $\displaystyle J/A$ is an element from $\displaystyle G/A$.

3. 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.

4. 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.

5. 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 $\displaystyle J$. If we take $\displaystyle x \in J$ then clearly $\displaystyle x \in G$ as $\displaystyle J \leq G$. Thus, $\displaystyle x^{-1}Ax = A$ as this holds for all $\displaystyle x \in G$.