Proof of Normal Subgroups in Group G - HN

[roam]

If H and N are subgroups of a group G. And we define $$HN = \{ hy | h \in H, y \in N \}$$,

Then I know that the following are true:

1. If $$N$$ is a normal subgroup. Then $$HN$$ is a subgroup of $$G$$.
   
2. If $$H$$ and $$N$$ are both normal subgroups. Then $$HN$$ is normal.

But does anybody know the proof to any of them?

For 2 I know that if they are both normal, then $$G=HN$$ and $$H \cap N = \{ e \}$$, but really I don't know how to prove it.

Any help or suggestions is appreciated.

 

[quasar987]

For 1., everything is obvious except the fact that HN is closed under the group operation, right? By definition, N normal means that for any g in G, $gNg^{-1}\subset N$. And notice that this is equivalent to $gN\subset Ng$. By using both these "equations", it is not hard to show that HN is closed under the group operation.

2. Think "multiplying by e in the form $g^{-1}g$". (This may help you also for 1.)

 

[roam]

Thanks. For (1), to show that it's a subgroup I could use the "one-step subgroup test".

First I have to show that HN is nonempty ($$HN \neq \emptyset$$). Then if h=e, hy=gg^-1y = g y g^-1 which is y. So is that okay?

Then assuming that two elements a and b have are in HN, using the assumption I have to show that ab^-1 is in HN.

So, to do that I take $$a \in H$$ and $$b \in N$$. But how would you represent b^-1?

 

[Office_Shredder]

Thanks. For (1), to show that it's a subgroup I could use the "one-step subgroup test".

First I have to show that HN is nonempty ($$HN \neq \emptyset$$). Then if h=e, hy=gg^-1y = g y g^-1 which is y. So is that okay?

Then assuming that two elements a and b have are in HN, using the assumption I have to show that ab^-1 is in HN.

So, to do that I take $$a \in H$$ and $$b \in N$$. But how would you represent b^-1?

Why does your one step subgroup test have two steps?

I'm really confused by how you show HN is non-empty. What are g and y supposed to be, and how do you know they commute? It seems like the answer is far more obvious: e is in H, e is in N, so ee=e is in HN. Therefore it's non-empty.

There's no real reason why a should be in H or b should be in N. All you have is an element a that's of the form hn and an element b that's of the form h'n' where h' and n' are elements of H and N respectively. Now what can you say about ab^-1? (it may help to calculate what b^-1 is in terms of h' and n')

 

[quasar987]

You took a in H and b in N. But this is not what the one-step subgroup test tells you to do. It tells you to take two arbitrary elements a and b in HN. So a is of the form a=hn for some h in H and n in N and b is of the form b=h'n' for some h' in H and n' in N. So

ab^{-1}=(hn)(h'n')^{-1}

and from there, manipulate this equation to get it into the form ab^{-1}=h''n'' for some h'' in H and n'' in N, thus showing that ab^{-1} is in HN.

 

[roam]

Thanks guys. Here's what I did for (1):

Suppose $$a=hy$$ and $$b=h'y', h,h' \in H$$ and $$y, y' \in N$$.

ab^-1=hy(h'y')^-1

By the "Socks-Shoes Property" we get:

=hyy'^-1h'^-1

(Since N is a normal subgroup of G we have

h'yy'^-1h'^-1 = z

That is yy'^-1h'^-1=h'^-1z, where $$z \in N$$.)

Hence: hyy'^-1h'^-1 = hh'^-1 z

For some z in N and hh'^-1 in H.

Right?

For part (2) I tried using the "normal subgroup test":

$$HN \triangleleft G \iff xHNx^{-1} \subseteq HN$$, $$\forall x \in G$$

xHN=H'N'x

xHNx^-1=H'N' (for some H'N' in HN)

So, $$xHNx^{-1} \subseteq HN$$

I could also show the converse of this. But I don't think think this proof is correct. Any help is appreciated.

 

[quasar987]

By the "Socks-Shoes Property" we get:

This is gold!

 

[quasar987]

Thanks guys. Here's what I did for (1):

Suppose $$a=hy$$ and $$b=h'y', h,h' \in H$$ and $$y, y' \in N$$.

ab^-1=hy(h'y')^-1

By the "Socks-Shoes Property" we get:

=hyy'^-1h'^-1

(Since N is a normal subgroup of G we have

h'yy'^-1h'^-1 = z

This is not true. Normal means hNh^-1<N for all h, but this does not imply that hNh'^-1 for h different from h'.

Hint: This is where you should try squeezing in an e=gg^-1 somewhere in there.

For part (2) I tried using the "normal subgroup test":

$$HN \triangleleft G \iff xHNx^{-1} \subseteq HN$$, $$\forall x \in G$$

xHN=H'N'x

xHNx^-1=H'N' (for some H'N' in HN)

So, $$xHNx^{-1} \subseteq HN$$

I could also show the converse of this. But I don't think think this proof is correct. Any help is appreciated.

First of all, don't use HN (or H'N') to denote an element of the subgroup HN. Capitals denote groups. Little letters denote elements of those groups. So here you consider an arbitrary element g in G and you must show that gHNg^-1<HN. Meaning that for any hn in HN, there exists h'n' in HN such that ghng^-1=h'n'. What could h' and n' be, knowing that H and N are normal? Again, think multiplication by e.

 

[roam]

This is not true. Normal means hNh^-1<N for all h, but this does not imply that hNh'^-1 for h different from h'.

Hint: This is where you should try squeezing in an e=gg^-1 somewhere in there.

It is possible that h'=h but it doesn't have to be. You have mentioned multiplication by e before, could you please show me clearly what you mean? Because I have no idea...

First of all, don't use HN (or H'N') to denote an element of the subgroup HN. Capitals denote groups. Little letters denote elements of those groups. So here you consider an arbitrary element g in G and you must show that gHNg^-1<HN. Meaning that for any hn in HN, there exists h'n' in HN such that ghng^-1=h'n'. What could h' and n' be, knowing that H and N are normal? Again, think multiplication by e.

for $$g \in G$$, gHN is a coset of the form:

$$gHN = \{ ghn | h \in H, n \in N \}$$

since H is normal gh can be written as h'g:

$$gHN = \{ h'gn | h \in H, n \in N \}$$

Also since $$N \triangleleft G$$, we can write gn=n'g

$$gHN = \{ h' n' g | h \in H, n \in N \} = HNg$$

Is this enough to show that HN is normal?

 

[quasar987]

It is possible that h'=h but it doesn't have to be. You have mentioned multiplication by e before, could you please show me clearly what you mean? Because I have no idea...

Fair enough if you've never seen this trick before. See below.

for $$g \in G$$, gHN is a coset of the form:

$$gHN = \{ ghn | h \in H, n \in N \}$$

since H is normal gh can be written as h'g:

$$gHN = \{ h'gn | h \in H, n \in N \}$$

Also since $$N \triangleleft G$$, we can write gn=n'g

$$gHN = \{ h' n' g | h \in H, n \in N \} = HNg$$

Is this enough to show that HN is normal?

That works, yes. (But there are typos: you need to change h for h' and n for n' in those sets)

Here is another proof. I am showing you this so that you see what I mean by "multiplying by e=g^-1g".

Consider an arbitrary element g in G. We must show that gHNg^-1<HN. Meaning that for any hn in HN, $ghng^{-1}\in HN$. Well, notice that e, the identity element of the group G can be written as e=g^-1g. So we have ghng^-1=gheng^-1=gh(g^-1g)ng^-1=(ghg^-1)(gng^-1). The first factor belongs to H, since H is normal, and the second factor belongs to N, since N is normal. Thus, the product belongs to HN, QED.

So the trick was to write e as g^-1g and then squeeze it at the right place. In this case, between the h and the n.
I'm saying that you can make progress in problem (1) by using the same trick.

 

[roam]

Thank you very much, I understand it now.

And sorry about the typos, I meant:

$$gHN = \{ ghn | h \in H, n \in N \}$$

$$gHN = \{ h'gn | h' \in H, n \in N \}$$

and $$gHN = \{ h' n' g | h' \in H, n' \in N \} = HNg$$.

 

[roam]

P.S. where would you sqeeze in the e=gg^-1 in (1), considering that g and g' are not usually the same and gg'¹=e iff g=g'?

 

[quasar987]

ab^-1=hy(h'y')^-1

By the "Socks-Shoes Property" we get:

=hyy'^-1h'^-1

=heyy'^-1h'^-1

I leave to you to find how to express e in this case, and to complete the argument.