# If P is normal in H, and H normal in G, then P is normal in G

• mathgirl313
In summary, in this conversation, the participants discuss the properties of finite groups and Sylow P-Subgroups. They focus on showing that if P is normal in H and H is normal in G, then P is normal in G. They discuss the use of conjugates and the uniqueness of P as a Sylow P-Subgroup in G. They ultimately reach the conclusion that P and gPg^-1 are both Sylow P-Subgroups of H and are therefore conjugate. Since P is normal in H, this means that gPg^-1 must be equal to P, and therefore P is normal in G.
mathgirl313

## Homework Statement

Now, G is a finite group, and P is a Sylow P-Subgroup of G, and H is a subgroup of G with P≤H≤G. So if P is normal in H, and H is normal in G, then P is normal in G.

## The Attempt at a Solution

I know I want to show that gpg^-1 is contained in P but that's about as far as I can get. I know gPg^-1 is going to be another Sylow P-Subgroup, but other than that I'm kind of stuck. When I try going about showing that gpg^-1 q, where q is an element in p, I have trouble showing the two elements are in P. I get as far as q=hph^-1 since q is in P and i think my best bet for showing it is contained in P is for closure...but that's all I get. I know I need to use the P is Sylow P-Subgroup since the property normally isn't transitive, but I'm not sure how to.

Thanks for any help.

All sylow p subgroups are conjugate, right?

Ya, gPg^-1 is going to be a conjugate to P, but why does that help me show that it is P? Couldn't it be another Sylow P-Subgroup?

mathgirl313 said:
Ya, gPg^-1 is going to be a conjugate to P, but why does that help me show that it is P? Couldn't it be another Sylow P-Subgroup?

If it's another Sylow P-Subgroup then that subgroup is contained in H. What next?

Dick said:
If it's another Sylow P-Subgroup then that subgroup is contained in H. What next?

Why is it necessarily going to be contained in H? Just because of orders and conjugate? Are you going to get that P is unique and so it's normal?

mathgirl313 said:
Why is it necessarily going to be contained in H? Just because of orders and conjugate? Are you going to get that P is unique and so it's normal?

Yes, I am. gHg^(-1)=H and P is contained in H.

Dick said:
Yes, I am. gHg^(-1)=H and P is contained in H.

So is gPg^(-1)=P because of that?

I'm sorry, I've always been dreadful at showing that something is a normal subgroup.

mathgirl313 said:
So is gPg^(-1)=P because of that?

I'm sorry, I've always been dreadful at showing that something is a normal subgroup.

No, that's not the reason. P and gPg^(-1) are two p-sylow subgroups of H. Think about that.

Dick said:
No, that's not the reason. P and gPg^(-1) are two p-sylow subgroups of H. Think about that.

Since they're normal in H they have to be the same subgroup.

mathgirl313 said:
Since they're normal in H they have to be the same subgroup.

There's a little more to it than that. P is normal in H. Why is gPg^(-1)?

Dick said:
There's a little more to it than that. P is normal in H. Why is gPg^(-1)?

Why is it normal? I can only think that since P is a Sylow P-Subgroup of H, and P is normal in H. But a p-Sylow group is normal if and only if the number of Sylow p-subgroups is one (for that
). So P is the unique Sylow p-subgroup. So P is the unique Sylow p-subgroup in G as well, and as such it's normal.

mathgirl313 said:
Why is it normal? I can only think that since P is a Sylow P-Subgroup of H, and P is normal in H. But a p-Sylow group is normal if and only if the number of Sylow p-subgroups is one (for that
). So P is the unique Sylow p-subgroup. So P is the unique Sylow p-subgroup in G as well, and as such it's normal.

Yeah, that does it. Since P and gPg^(-1) are both sylow p-subgroups of H. Then they are conjugate in H. So gPg^(-1)=hPh^(-1) where h is in H. Since P is normal in H, gPg^(-1)=P.

Dick said:
Yeah, that does it. Since P and gPg^(-1) are both sylow p-subgroups of H. Then they are conjugate in H. So gPg^(-1)=hPh^(-1) where h is in H. Since P is normal in H, gPg^(-1)=P.

Awesome! Thank you do much for walking me through it!

## 1. What does it mean for a subgroup to be normal in a group?

A subgroup H of a group G is normal if and only if for every element g in G and every element h in H, the conjugate ghg^-1 is also in H. In other words, the left and right cosets of H in G are equal, and the group operation is preserved when multiplying by elements in H.

## 2. How does normality of subgroups relate to the overall group structure?

The normality of a subgroup is an important property that allows us to study the structure of a group. It ensures that certain operations, such as taking quotient groups, are well-defined. Additionally, normal subgroups are central to the concept of a group homomorphism, which is a key tool in understanding the relationship between different groups.

## 3. Can a subgroup be normal in more than one group?

Yes, a subgroup can be normal in multiple groups. This is because the definition of normality depends on the specific group in which the subgroup is being considered. If a subgroup satisfies the criteria for normality in multiple groups, then it is considered normal in each of those groups.

## 4. How does the normality of subgroups affect the simplicity of a group?

The normality of subgroups plays a crucial role in the simplicity of a group. A group is considered simple if it has no non-trivial normal subgroups. This means that the only normal subgroups of a simple group are the trivial subgroup and the entire group itself. Therefore, if a subgroup is normal in a simple group, then it must be either trivial or the entire group.

## 5. Is the converse of the statement "If P is normal in H, and H normal in G, then P is normal in G" true?

No, the converse of this statement is not necessarily true. Just because a subgroup P is normal in a subgroup H, and H is normal in a group G, does not mean that P is normal in G. This is because normality is not a transitive property in general. However, if P is normal in H and H is normal in G, and P is a normal subgroup of G, then the statement is true.

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