# Homework Help: Proving a Definition ?

1. Nov 23, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
A subgroup $H$ of a group $G$ is normal iff $\forall g \in G$ and $\forall h \in H$, $ghg^{-1} \in H$.

2. Relevant equations

3. The attempt at a solution
I am having trouble seeing how this is not saying the same thing as the definition of a normal subgroup. Could somebody expound the difference between this statement and the definition of a normal subgroup?

2. Nov 23, 2014

### Fredrik

Staff Emeritus
Your goal is to prove that the following two statements are equivalent:

(a) For all $g\in G$, we have $gH=Hg$.
(b) For all $g\in G$ and all $h\in H$, we have $ghg^{-1}\in H$.

Edit: I also have to add that you never prove a definition.

Last edited: Nov 24, 2014
3. Nov 23, 2014

### Orodruin

Staff Emeritus
Since you have been tasked with showing this, there should not be a difference and you could just as well choose this as the definition. That being said, the proof should be trivial as you say.

4. Nov 26, 2014

### Bashyboy

Yes, I am aware of that, which is why I included a question mark in the subject of this thread; the way the problem was stated made it appear that I was to prove a definition.

So, we are trying to verify the statement $\forall g \in G$, $gH = Hg$ iff $\forall h \in H$, every element can be written as $ghg^{-1}$ (i.e., $ghg^{-1} \in H$).

First implication: Suppose that $gH = Hg$ is true. By definition, this means that $H$ is normal, and that every element in $H$ can be written in the form $ghg^{-1}$, where $h \in H$.

Second implication: Suppose that $\forall h \in H$, $g h g^{-1}$ is also in $H$. Then every element in $H$ an be written as $ghg^{-1}$, which means that $H$ is normal.

Does this proof appear correct?

5. Nov 26, 2014

### Fredrik

Staff Emeritus
You skipped all the steps of the proof between the assumption and the conclusion. You have to make it clear that you're using the assumptions and the definitions, and also how you're using them.

First implication: Suppose that $gH=Hg$ for all $g\in G$. Let $g\in G$ and $h\in H$ be arbitrary. We have $gh\in gH=Hg$. This implies that...

6. Nov 26, 2014

### pasmith

You haven't proven anything.

Consider the following statements:

Statement 1
x is "good" if and only if x has property P.

Statement 2
x is "good" if and only if x has property Q.

These are only consistent if "x has property P" and "x has property Q" are equivalent. But that has to be shown. The consequence is that exactly one of those statements can be labelled a "definition" (and thus assumed to be true), and the other must be labelled a "proposition" or "theorem" (and must be proven to be true).

Your definition of "normal" is that H is normal if and only if for all $g \in G$ we have $gH = Hg$. Your task is to prove that the statement "for all $g \in G$ we have $gH = Hg$" is equivalent to the statement "for all $g \in G$ and all $h \in H$ we have $ghg^{-1} \in H$". Having done that, you will have established the proposition that H is normal if and only if for all $g \in G$ and all $h \in H$ we have $ghg^{-1} \in H$.

Suppose then that H is such that for all $g \in G$, $gH = Hg$. Hence if $h_1 \in H$ there exists an $h_2 \in H$ such that $gh_1 =$...

7. Nov 26, 2014

### Bashyboy

I suppose I am still having difficulty discerning the difference between the statement given in the first post and the definition of normal subgroup. Here is the definition of a normal subgroup:

A subgroup $H$ of a group $G$ is normal if and only if for every $g \in G$, $gH = Hg$, or, equivalently, $H = gHg^{-1}$. (The final set is shorthand for $gHg^{-1} = \{ghg^{-1} ~|~h \in H\}$).

Last edited: Nov 26, 2014
8. Nov 26, 2014

### Fredrik

Staff Emeritus
This definition only makes sense if you have already proved that the two statements are equivalent. Forget about the term "normal" for now, and just focus on proving that the two statements are equivalent, preferably without using the word "normal" anywhere in the proof.

9. Nov 26, 2014

### Bashyboy

But this definition precedes the theorem which I am currently trying to prove. So, how could the definition only make sense if I first proved this theorem?

10. Nov 26, 2014

### Fredrik

Staff Emeritus
I don't know what exactly your book is doing, but there's a number of approaches you can take here. You can define "normal" by saying that H is normal if it satisfies condition (a) (in post #2), and then prove that a subgroup H is normal if and only if it satisfies condition (b). Alternatively, you can define "normal" by saying that H is normal if it satisfies condition (b), and then prove that a subgroup H is normal if and only if it satisfies condition (a). A third approach is to prove that (a) and (b) are equivalent, and then define "normal" by saying that H is normal if it satisfies (a) and (b). If the book uses a version of that last approach in which the definition is stated before the theorem, then the author is doing something weird, but not really wrong, since he didn't make any false statements. The only problem is that the reader won't know that the definition makes sense until he has proved the theorem.

Note that each of these approaches ensures that the statements (a) and (b) are true if and only if H is normal.

Edit: Based on your statement of the problem in post #1, I'd say that your book is using the first approach I described above. Definition: H is normal if (a). Theorem: (a) is equivalent to (b).

Last edited: Nov 26, 2014
11. Nov 27, 2014

### Bashyboy

Okay, well I am still uncertain as to what is wrong with what I expressed in post number four. Here is another attempt to prove the conditional statement "If $gH = Hg$, then \forall h \in H$, ghg^{-1} \in H$.

Suppose that $gH = Hg$ is true. Let $x \in gH$ be arbitrary. If $x$ is in this set, then it can be written in the form $x = gh_1$, where $h_1$ is some element in $H$. Furthermore, because $x \in gH$, it must also be in $gH$, that is, $x = h_2 g$. Equating the two,

$gh_1 = h_2 g \iff$

$gh_1 g^{-1}_2 = h_2$.

Since $h_2$ is just some element in $H$, and we have shown that it can be written in the form $g h g^{-1}$, then $gH = Hg$ being true implies that every element in $H$ can be written in the form $ghg^{-1}$.

Would this suffice for the first implication?

12. Nov 27, 2014

### Fredrik

Staff Emeritus
Yes, that's pretty good. However, we didn't want to prove that each element of H can be written in the form $ghg^{-1}$. (That would be the statement "For all $h\in H$, there's a $g\in G$ and a $h'\in H$ such that $h=gh'g^{-1}$.") What we want to prove at that point is the statement (b) from post #2. Since you proved enough to make it obvious that (b) holds, I would say that your proof is OK, but the end of it can be done more clearly.

This is my proof of the implication $(a)\Rightarrow (b)$:

Suppose that $gH=Hg$ for all $g\in G$. Let $g\in G$ and $h\in H$ be arbitrary. We have $gh\in gH=Hg$. Let $h'\in H$ be such that $gh=h'g$. (Such a $h'$ exists because $gh\in Hg$). We have $ghg^{-1}=h'\in H$. Since $g$ is an arbitrary element of $G$ and $h$ is an arbitrary element of $H$, this means that (b) holds.

13. Nov 27, 2014

### Bashyboy

Okay, now for the second implication: "If every element in $H$ can be written as $ghg^{-1}$, then $H$ is normal."

Doesn't the truthfulness of this implication immediately follow from the definition of normal, which states that every element in $H$ is of the form $gh^{-1}$?

14. Nov 27, 2014

### Fredrik

Staff Emeritus
What definition of "normal" are you using now? I'm also confused about why you changed the original statement (corresponding to my (b)) from post #1 to post #3. The original statement can be worded like this:

For all $g\in G$ and all $h\in H$, we have $ghg^{-1}\in H$.​

I'm not even sure what you mean by the new statement. Do you mean

For all $g\in G$, every element of $H$ can be written as $ghg^{-1}$ with $h\in H$.​

or

Every element of $H$ can be written as $ghg^{-1}$ with $h\in H$ and $g\in G$.​

?

I guess it's not the latter, because that version of the statement is proved by the trivial $h=he=hhh^{-1}$. So I assume it's the former. I'll call that statement (c). It's fairly easy to prove that $(b)\Leftrightarrow (c)$, but I don't understand why you changed from (b) to (c) between posts #1 and #3. (I didn't notice that you did until now).

Was post #11 an attempt to prove $(a)\Rightarrow (b)$ or $(a)\Rightarrow (c)$? My response was based on the assumption that you wanted to prove $(a)\Rightarrow (b)$. If you meant to prove $(a)\Rightarrow (c)$, then I don't think you succeeded. You proved that $h_2$ can be written in the form $ghg^{-1}$ with $h\in H$, but $h_2$ wasn't defined as an arbitrary element of $H$. It was defined as the specific $k\in H$ such that $x=kg$.

15. Nov 27, 2014

### Bashyboy

I am using the definition of normal provided in post number 7.

I am just trying to prove that the two statements, which you gave in post number two, are equivalent.

16. Nov 27, 2014

### Bashyboy

I am using the definition of normal provided in post number 7.

17. Nov 27, 2014

### Fredrik

Staff Emeritus
You seem to have chosen to prove that these two statements are equivalent:

(d) H is normal (in the sense of post #7)
(e) For all $g\in G$, we have $H\subseteq gHg^{-1}$.

That doesn't make any sense. If you want to prove that (d) implies (e), then the fact that (d) relies on the meaning of "normal", you're forced to use the definition in post #7, which says explicitly that (d) implies that $H=gHg^{-1}$ for all $g\in H$, which of course implies (e). So your own definition forces you to use what you're trying to prove to prove what you're trying to prove.

What you need to do to make sense of the definition in #7 is to prove that the following two statements about an arbitrary $g\in G$ are equivalent:

(f) $gH=Hg$.
(g) $H=gHg^{-1}$

Just start with "Let $g\in G$ be arbitrary, and then prove that (f) implies (g) and that (g) implies (f). Then you will know that the definition in #7 makes sense.

18. Nov 28, 2014

### vela

Staff Emeritus
The author of your book did something weird in defining what a normal subgroup is by giving you two possible definitions:
1. For all $g\in G$, we have $gH=Hg$.
2. For all $g\in G$ and all $h\in H$, we have $ghg^{-1}\in H$.
and asserting that they are equivalent. The problem is asking you to show that this assertion is justified.

To put it another way, the problem is asking you to prove the author's definition is valid. Before you prove anything, it is certainly possible that statement (1) and statement (2) are inconsistent with each other, and the definition isn't well conceived, right? Because you're trying to prove the definition makes sense, you can't assume it's true. That was the problem with your attempt in post 4. If you assume the definition is valid, you can indeed simply say (2) -> normal subgroup -> (1), and (1) -> normal subgroup -> (2). But it's not a valid proof that (1) <-> (2) because the definition assumes (1) <-> (2).

Fredrik's asking you to ignore the author's definition altogether and simply look at the two statements by themselves and show that each statement implies the other. Once you've established that, then you can accept that the author's definition isn't wacky.

Finally, to repeat what Fredrik said a few posts ago, the problem appears to be using statement (1) as the sole definition of a normal subgroup and not statement (2). This is inconsistent with the way the author defined a normal subgroup, so it's no surprise that you found the problem confusing.

19. Nov 28, 2014

### Fredrik

Staff Emeritus
At the start of the thread, I didn't realize that there are at least two separate issues. I assumed that your definition of "normal" was just that H is said to be normal if $gH=Hg$ for all g. But you were using the following definition:
And you wanted to use it to solve the following problem:

If you want to verify that the definition of "normal" quoted above makes sense, then you have to prove that the equivalence $gH=Hg\Leftrightarrow H=gHg^{-1}$ holds for all $g\in G$. Once you have done that, you can attack the problem with more confidence. If we use the definition of "normal" quoted above, then this is the strategy:

Suppose that H is normal. Let $g\in G$ and $h\in H$ be arbitrary. Can you prove that $ghg^{-1}\in H$?

Conversely, suppose that for all $g\in G$ and all $h\in H$, we have $ghg^{-1}\in H$. Let $g\in G$ be arbitrary. Can you prove that H is normal, e.g. by proving that $gHg^{-1}=H$?

20. Nov 30, 2014

### Bashyboy

Okay, I have been told by my professor that the problem translates to:

$H$ is normal iff $gHg^{-1} \subseteq H$ $\forall g \in G$.

First implication: If $H$ is normal, then $gHg^{-1} \subseteq H$.

Suppose $H$ is normal. By definition, this means that $H = gHg^{-1}$. And so, we can justly conclude that $gHg^{-1} \subseteq H$.

I need to prove one idea before proving the second implication: If $x \in gHg^{-1}$, then $x^{-1} \in gHg^{-1}$.

Let $x \in gHg^{-1}$ be arbitrary. Then $x = ghg^{-1}$. Taking the inverse, we get

$x^{-1} = [g(hg^{-1})]^{-1} \iff$

$x^{-1} = (hg^{-1})^{-1}g^{-1} \iff$

$x^{-1} = gh^{-1}g^{-1}$

Because $gh^{-1} g^{-1} \in gHg^{-1}$, then $x^{-1}$ must also be in this set.

Second implication: If $gHg^{-1} \subseteq H$, $\forall g \in G$, then $H$ is normal (that is, $H = gHg^{-1}$)
.
.
.
Hold on...My proof of the second implication is wrong...I am out of ideas...

Last edited: Nov 30, 2014
21. Nov 30, 2014

### vela

Staff Emeritus
Perhaps you should consider $g^{-1}hg$ rather than finding the inverse of $ghg^{-1}$.

22. Dec 1, 2014

### Fredrik

Staff Emeritus
To prove the second implication, it's sufficient to prove that $H\subseteq gHg^{-1}$ for all $g\in G$. So let $g\in G$ and $h\in H$ be arbitrary. Now try to prove that $h\in gHg^{-1}$.

23. Dec 4, 2014

### Bashyboy

Fredik, I am most certain that the second implication is to prove that $gHg^{-1} \subseteq H$, $\forall g \in H$ and $\forall h \in H$; but I may be wrong.

This is what I am having prodigious difficulty with. I am trying to take the arbitrary element $ghg^{-1}$ of the set $gHg^{-1}$ and demonstrate that this element in is $H$.

24. Dec 4, 2014

### vela

Staff Emeritus
The second implication is the opposite direction: if $gHg^{-1} \subseteq H$ for all $g \in G$, then $H$ is normal.

25. Dec 4, 2014

### Bashyboy

Vela: Yes, you are right. I forgot to include the part about $H$ being normal is a consequence of $gHg^{-1} \subseteq H$--which I am having difficulty with. I am having trouble showing that $ghg^{-1}$ is also an element of $H$.

Last edited: Dec 4, 2014