# Normal Subgroup of G

1. Jul 31, 2014

### Justabeginner

1. The problem statement, all variables and given/known data
N is a normal subgroup of G if aNa^-1 is a subset of N for all elements a contained in G. Prove that in that case aNa^-1 = N.

2. Relevant equations

3. The attempt at a solution
Given: N is a normal subgroup of G if aNa^-1 is a subset of N for all elements a contained in G. Assume, aNa^-1 = {ana^-1|element n in N}.
By definition of a normal subgroup, aN is a subgroup of Na for all elements a in G. Then aNa^-1 is a subgroup of Naa^-1 and is = N. And a^-1Na is a subgroup of aNa^-1 and is = N for all elements a in G, when Na= a(a^-1N)a is a subgroup of a(Na^-1)a = aN.

Is this approach correct?

2. Jul 31, 2014

### jbunniii

No, $aN$ and $Na$ are not groups at all unless $a \in N$. If $N \unlhd G$ then it is true that $aN = Na$ (i.e. there is no distinction between left and right cosets). But this is not the usual definition of a normal subgroup.

Indeed, the definition given in your problem statement is that $N \unlhd G$ if and only if $aNa^{-1} \subseteq N$ for all $a \in G$. The goal is to prove that this implies that $aNa^{-1} = N$ for all $a \in G$. You can do this without mentioning cosets at all.

Hint: $aNa^{-1} \subseteq N$ if and only if $N \subseteq a^{-1}Na$.

3. Jul 31, 2014

### Justabeginner

So if, aNa^-1 is a subset of N if and only if N is a subset of a^-1Na, then element a is contained in N. Then, aNa^-1= N for all elements a in G.

4. Jul 31, 2014

### jbunniii

No, you cannot conclude that $a \in N$. Indeed, $a$ is an arbitrary element of $G$.

5. Jul 31, 2014

### Justabeginner

So if my conclusion is that elements a are present in G, and aNa^-1 = N for all elements a in G, this proves that N is a normal subgroup of G.

(I realize how I cannot say that element a is contained in N now).

6. Jul 31, 2014

### jbunniii

The definition in your problem statement is that $N \unlhd G$ provided that $aNa^{-1} \subseteq N$ for all $a\in G$.

The goal is to conclude the apparently stronger statement that if $N \unlhd G$, then in fact we must have equality: $aNa^{-1} = N$ for all $a \in G$. Since this is an equality of sets, the usual way to prove it is to prove containment in both directions, namely $aNa^{-1} \subseteq N$ and $N \subseteq aNa^{-1}$. The first containment is given (definition of normality), so you just need to prove the second one.

Try thinking about my hint some more: $aNa^{-1} \subseteq N$ if and only if $N \subseteq a^{-1}Na$

7. Jul 31, 2014

### micromass

To give a further hint. If you want to prove an inclusion such as $N\subseteq aNa^{-1}$, you should always do the following; pick an arbitrary element $x\in N$ (so the only thing you know about $x$ is that it is in $N$). You must proe that $x\in aNa^{-1}$. In other words, you must find some $y\in N$ such that $x=aya^{-1}$.

(Yes, jbunniii, I know this is not the approach you had in mind, but I think this seems simpler)

8. Jul 31, 2014

### jbunniii

No problem, any proof is fine as long as it correct and (most importantly) Justabeginner understands it. I don't care if it's my proof or not.

9. Jul 31, 2014

### micromass

Your method is still important though, so be sure to give it after he finds the proof himself.

10. Jul 31, 2014

### Justabeginner

Thank you both for the techniques.

So suppose element x is in N, where x=aya^-1, then aya^-1 is contained in N. So, y is contained in a^-1Na and therefore if a^-1xa is contained in a^-1Na as well, x is contained in a^-1Na?

11. Jul 31, 2014

### jbunniii

How do you know that you can write $x$ in the form $aya^{-1}$? It is true, but you haven't shown why.

May I suggest instead starting as follows: $x \in N$, and $N$ is a normal subgroup. Therefore, given any $g \in G$, what can you say about $gxg^{-1}$?

12. Jul 31, 2014

### Justabeginner

If element x is in N, and N is a normal subgroup of G, given any element g in G, gxg^-1 must also be in G.

Edit: gxg^-1 has to be in N too, because x is in N.

Last edited: Jul 31, 2014
13. Jul 31, 2014

### micromass

Good. So what if we know define $y$ as $axa^{-1}$?

14. Jul 31, 2014

### jbunniii

Small correction: "must also be in N."

15. Jul 31, 2014

### Justabeginner

Now, if element y is contained in N, and y= axa^-1 then axa^-1 is also contained in N. If axa^-1 is in N and N is a subgroup of G, then axa^-1 is also in G, so a is in G.

16. Jul 31, 2014

### micromass

So we have the following:

1) We know that $y = axa^{-1}$ by definition.
2) We know that $aNa^{-1} \subseteq N$ since $N$ is normal by hypothesis.
3) We know that $x\in N$

Can you use this to establish the following facts:

4) $x = aya^{-1}$
5) $y\in N$

Finally, can you use those 5 facts to establish that $x\in aNa^{-1}$?

17. Jul 31, 2014

### Justabeginner

By multiplication, x=a^-1ya. Since x is in N, and x=a^-1ya, y is also in N. If y is in N, then a^-1xa is in N, and x is in aNa^-1.

Last edited: Jul 31, 2014
18. Jul 31, 2014

### jbunniii

Shouldn't that be $x = a^{-1}ya$?

19. Jul 31, 2014

### micromass

Yes, sorry for the confusion.

20. Jul 31, 2014

### Justabeginner

I forgot to put them on the correct side when I multiplied.
Is the flow of my statements right?