# Thread: Normal Subgroups of S4

1. ## Normal Subgroups of S4

Prove that $\{ e,(12),(34),(12)(34) \}$ is a normal subgroup of $S_4$.

The only method really available to me at this point in my class is show that $xHx^{-1} \subset H$.

But $S_4$ has 24 elements and this would take a long time to check. Is there a more elegant solution?

Bear in mind. I only know of Normal Subgroups, Factors Groups and the internal direct product. So, no fancy conjugacy/sylow arguments please.

2. Originally Posted by Haven
Prove that $\{ e,(12),(34),(12)(34) \}$ is a normal subgroup of $S_4$.

The only method really available to me at this point in my class is show that $xHx^{-1} \subset H$.

But $S_4$ has 24 elements and this would take a long time to check. Is there a more elegant solution?

Bear in mind. I only know of Normal Subgroups, Factors Groups and the internal direct product. So, no fancy conjugacy/sylow arguments please.
Have you encountered the first isomorphism theorem yet? If so, you should note that it says that kernels of homomorphisms are precisely the same thing as normal subgroups.

So, if you can find a homomorphism with this subgroup as a kernel then you are done. (HINT: you know the image will have 6 elements. How many groups of order 6 do you know?)

3. Unfortunately no, I don't know that theorem.

4. Originally Posted by Haven
Unfortunately no, I don't know that theorem.

I can't understand how you, according to yourself, know about factor groups but not the so basic first isomorphism theorem. You see, you make it very hard to help you since you reject practically every known system to show that subgroup is normal! Except, of course, to show directly that $xHx^{-1}\subset H\,\,\,\forall x\in S_n$ , which I'm afraid is what you're going to have to do.

Tonio

5. Originally Posted by tonio
I can't understand how you, according to yourself, know about factor groups but not the so basic first isomorphism theorem. You see, you make it very hard to help you since you reject practically every known system to show that subgroup is normal! Except, of course, to show directly that $xHx^{-1}\subset H\,\,\,\forall x\in S_n$ , which I'm afraid is what you're going to have to do.

Tonio
We're going through the book Contemporary Abstract Algebra by Joesph Gaillan. He doesn't even mention the first isomorphism theorem until his discussion on Sylow's theorems. In the section we're in, we have just defined some elementary properties of factor groups and normal groups.

6. @ Haven
Are you familiar with the following result?

Given groups $G$ and $H,$ if $\phi:G\to H$ is a homomorphism with kernel $K$ and is onto, then $G/K\cong H.$

This is the theorem that Swlabr is suggesting you should use to solve your problem.

@ Swlabr, tonio
The term “first isomorphism theorem” is not a standard one in the literature of group theory. Different books call the result above by different names. In A Course in Group Theory by John F. Humphreys, for example, the author refers to this result as the Homomorphism Theorem (reserving the term First Isomorphism Theorem for a different result). Most likely Haven’s book does not call the theorem above the First Isomorphism Theorem either. Instead of beating about the bush, it might have been more helpful to Haven if you had just stated the theorem explicitly, as I have done.

7. Originally Posted by proscientia
@ Haven
Are you familiar with the following result?
Given groups $G$ and $H,$ if $\phi:G\to H$ is a homomorphism with kernel $K$ and is onto, then $G/K\cong H.$
This is the theorem that Swlabr is suggesting you should use to solve your problem.

@ Swlabr, tonio
The term “first isomorphism theorem” is not a standard one in the literature of group theory. Different books call the result above by different names. In A Course in Group Theory by John F. Humphreys, for example, the author refers to this result as the Homomorphism Theorem (reserving the term First Isomorphism Theorem for a different result). Most likely Haven’s book does not call the theorem above the First Isomorphism Theorem either. Instead of beating about the bush, it might have been more helpful to Haven if you had just stated the theorem explicitly, as I have done.

Well, Gallian (whose book is the OP following) calls the First Isomorphism Theorem to what most algebraist call it. Humphreys seems to be of the very few that calls it "homomorphism theorem". Sometimes I've seen this theorem being called "isomorphism theorem", but never "homomorphism theorem".
Anyway Gallian has it with this name but, for some reason, he puts it way after factor groups, direct products, automorphism groups (!) and other stuff. Perhaps there's a good reason for this but it's a rather unorthodox order of subjects.

Finally, and whatever it be: it seems there is no much choice for the OP but to prove normality the long way since he's not yet covered other topics that'd help him to do it shorter.

Tonio

8. Well, I have news for you. We have
$(12)\in H$
but
$(13)(12)(13)^{-1}=(23)\notin H$
Hence $H=\{e,(12),(34),(12)(34)\}$ is not a normal subgroup of $S_4.$

Everyone happy now?

9. Originally Posted by proscientia
Well, I have news for you. We have
$(12)\in H$
but
$(13)(12)(13)^{-1}=(23)\notin H$
Hence $H=\{e,(12),(34),(12)(34)\}$ is not a normal subgroup of $S_4.$

Everyone happy now?

I think several of us knew that already. The problem was how to explain this to the OP without having him/her doing what you did for him/her: to explicitly try $xHx^{-1}\,,\,\,x\in S_n$ ...

Tonio

10. Originally Posted by tonio
I think several of us knew that already. The problem was how to explain this to the OP without having him/her doing what you did for him/her: to explicitly try $xHx^{-1}\,,\,\,x\in S_n$ ...

Tonio
I didn't, and my method wouldn't have picked up on this.

Ah well. One day I'll learn to think critically...

11. Originally Posted by Swlabr
I didn't, and my method wouldn't have picked up on this.

Ah well. One day I'll learn to think critically...

Unless you haven't had yet a basic course in group theory I wonder how you didn't know: perhaps you forgot some basic facts about permutation groups? For example, two permutations are conjugate iff they have exactly the same cycle decomposition.
From here it's plain that $(12)$ is conjugate to $(13),(14)$ and etc., so the subgroup cannot be normal if all these elements aren't contained in the subgroup...

Tonio