# Two Problems In Group Theory

• Feb 25th 2008, 11:10 AM
notyeteuler
Two Problems In Group Theory
Hello there :)

Wow, this forum is amazing! I didn't realize places like this exist and I'm really looking forward to getting stuck in to helping people. Anyway, my first post is unfortunately pleading for help with a subject that I have very little in common with. However, I have an upcoming assessment and am really struggling. Any help that you could give me would be very much appreciated

1/ Let G be a finite group and N be a normal subgroup of G such that gcd(|N|,|G/N|)=1. Show that N is the only subgroup of G with order |N|.
I have the feeling that this has something to do with the Second Isomorphism Theorem, though I am not sure. I have tried everything I can think of involving it but to no avail

2/ Give a decomposition series for the dihedral group of order 28. To which well-known groups are the composition factors isomorphic? I'm lost on this one :(

Thank you in advance for any help that you can give

Paul
• Feb 25th 2008, 11:39 AM
ThePerfectHacker
Quote:

Originally Posted by notyeteuler
1/ Let G be a finite group and N be a normal subgroup of G such that gcd(|N|,|G/N|)=1. Show that N is the only subgroup of G with order |N|.
I have the feeling that this has something to do with the Second Isomorphism Theorem, though I am not sure. I have tried everything I can think of involving it but to no avail

I have to go soon. I did not solve it yet, but I can solve if for abelian groups, not sure if this is the approach you are looking for. Suppose that $|G|>1$ then $|G| = p_1^{a_1}\cdots p_k^{a_k}$. If $H\subseteq G$ such that $\gcd(|H|,|G/H|)=1$ it means $H=p_1^{b_1}\cdots p_k^{b_k}$ where $b_i = a_i \mbox{ or }0$. Suppose then $H=p_1^{a_1}\cdots p_l^{a_l}$ let $P_1,...,P_l$ be the $p_1,...,p_l$ Sylow subgroups. Then $P_1...P_l = H$ because there can be only one Sylow subgroup since the group is abelian all of them are normal. (What an ugly proof :().
• Feb 25th 2008, 11:42 AM
notyeteuler
Thank you for your help :) However, we I have not yet learned about Sylow subgroups and expect that we are not meant to use them :(
• Feb 25th 2008, 06:14 PM
ThePerfectHacker
We can prove something stronger. :)

Theorem: Let $G$ be a finite group and $N$ a normal subgroup with $\gcd( |N|,|G/N|) = 1$ and if $H\subseteq G$ is a subgroup with $(|H|) \big| (|N|)$ then $H\subseteq N$.

Proof: Form the factor group $G/N$. Let $a\in H$ and consider the coset $aN$, we will show that $aN=N$ proving that $a\in N$ and completing the proof. Note, it is sufficient to prove that the order of $aN$ in the group $G/N$ is $1$. Let $k$ be the order of $aN$. Let $n$ be the order of $a$ in the group $G$. Now $(aN)^n = a^nN = N$ thus it means $k|n$. But $k|(|G/N|)$ by Lagrange's theorem and $n$ divides $|H|$ so $n$ divides $|N|$ by Lagrange's theorem, thus $k$ divides $|N|$. But then $k=1$ because they are relatively prime.
• Feb 25th 2008, 10:18 PM
ThePerfectHacker
Quote:

Originally Posted by notyeteuler
2/ Give a decomposition series for the dihedral group of order 28. To which well-known groups are the composition factors isomorphic? I'm lost on this one :(

I assume you mean a composition series, there is no such thing as a decomposition series.

The dihedral group $\text{D}_{28}$ can be thought of $\{ 1,x,x^2,...,x^{13},xy,x^2y,...,x^{13}y|x^{14}=y^2= 1 \mbox{ and }yx=x^{13}y\}$.

Let $G_0 = \text{D}_{28}$.
Let $G_1 = \left< x \right> = \{1,x,x^2,...,x^{13}\}$.
Let $G_2 = \left< x^2 \right> = \{1,x^2,x^4,...,x^{12}\}$.
Let $G_3 = \{1\}$.

Now $G_3\triangleleft G_2\triangleleft G_1 \triangleleft G_0$ is a composition series. Because $G_0/G_1 \simeq \mathbb{Z}_2$ which is simple. And $G_1/G_2\simeq \mathbb{Z}_2$ which is simple. And $G_2/G_3\simeq \mathbb{Z}_7$ which is simple.

Note, all the composition factor groups are abelian which means this dihedral group is solvable. And furthermore, any composition series must has basically the composition series given above by the Jordan-Holder theorem (one of my favorite theorems from all math).
• Feb 25th 2008, 11:00 PM
notyeteuler
Thank you very much, that's amazingly helpful :)