Proving N is a Normal Subgroup of G

[Justabeginner]

Homework Statement

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.

Homework Equations

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?

 

[jbunniii]

Homework Statement

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.

Homework Equations

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.

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##.

 

[Justabeginner]

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##.

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.

 

[jbunniii]

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.

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

 

[Justabeginner]

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

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).

 

[jbunniii]

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.

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##

 

[micromass]

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##

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)

 

[jbunniii]

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

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.

 

[micromass]

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.

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

 

[Justabeginner]

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)

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?

 

[jbunniii]

Thank you both for the techniques.

So suppose element x is in N, where x=aya^-1

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}##?

 

[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.

 

[micromass]

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.

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

 

[jbunniii]

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.

Small correction: "must also be in N."

 

[Justabeginner]

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

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.

 

[micromass]

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.

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}##?

 

[Justabeginner]

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}##?

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.

 

[jbunniii]

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}##

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

 

[micromass]

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

Yes, sorry for the confusion.

 

[Justabeginner]

Yes, sorry for the confusion.

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