# Thread: Group Theory - Question on normal subgroup

1. ## Group Theory - Question on normal subgroup

This is from Herstein, q#6, section 2.9 pg 75

Question - order(G) = p^2, where p is a prime. Prove any normal subgroup, N, of order p, must lie in the center of G.

I have been trying this for a while, but stuck. This question comes much before concepts like class equation of G, Sylow's theorm etc. So argument used should be more basic. Any help/pointers to get me started plz?

I was trying to use the fact that N is cyclic and normal but was not successful.

2. Originally Posted by aman_cc
This is from Herstein, q#6, section 2.9 pg 75

Question - order(G) = p^2, where p is a prime. Prove any normal subgroup, N, of order p, must lie in the center of G.

I have been trying this for a while, but stuck. This question comes much before concepts like class equation of G, Sylow's theorm etc. So argument used should be more basic. Any help/pointers to get me started plz?

I was trying to use the fact that N is cyclic and normal but was not successful.
let $N=$ and $y \in G.$ if $o(y)=p^2,$ then $G$ would be cyclic, hence abelian, and we're done. also if $y=1_G,$ then $xy=yx.$ so we may assume that $o(y)=p.$ now we have $yxy^{-1}=x^j,$ for

some $j,$ because $N \lhd G.$ then $x=y^pxy^{-p}=x^{j^p}$ and therefore $x^{j^p - 1} = 1.$ hence $p \mid j^p -1,$ because $o(x)=p.$ but, by Fermat's little theorem, we also have that $p \mid j^p - j.$ thus $p \mid j - 1$ and

so $x^j = x.$ therefore $yxy^{-1}=x^j=x.$ hence $yx=xy. \ \Box$

Remark: a similar argument can be used to prove a more general result: if $G$ is any group of order $p^n,$ whre $p$ is prime, and $N \lhd G$ with $|N|=p,$ then $N \subseteq Z(G).$

3. Originally Posted by aman_cc
This is from Herstein, q#6, section 2.9 pg 75

Question - order(G) = p^2, where p is a prime. Prove any normal subgroup, N, of order p, must lie in the center of G.

I have been trying this for a while, but stuck. This question comes much before concepts like class equation of G, Sylow's theorm etc. So argument used should be more basic. Any help/pointers to get me started plz?

I was trying to use the fact that N is cyclic and normal but was not successful.
Can we not just apply the Burnside Basis Theorem and prove that all groups of order $p^2$ are Abelian? This would clearly prove the result.

Let $G$ be a finite $p$-group, and denote by $G'$ the derived subgroup, $G'=<[g,h]:g,h \in G>=$.

Denote by $\Phi(G)$ the Frattini subgroup; the intersection of all the maximal subgroups of $G$. Then $G/\Phi(G)$ is an elementary Abelian group which we can view as a vector space of rank $d(G)$. The Burnside Basis Theorem states:

i) $\Phi(G)=G'G^p$.

ii) For $S \subseteq G$ such that $=G$ then there exists $T \subseteq S$ such that $|T|=d(G)$ and $=G$.

iii) Another statement that isn't really relevant here but basically says that the generators of the group are precisely the coset representatives for the generators of the quotient group.

Thus, if $|\Phi(G)|=p$ we have that the group is cyclic (by part (ii) ), and if $|\Phi(G)|=1$ then we have that $G'=1$ and the group is Abelian (by part (i) ). Clearly there exists a subgroup of order $p$ in the group so the group cannot have $\Phi(G)=G$. Thus the group is Abelian.

I know it is a bit like overkill, but it is my favorite theorem...

4. Originally Posted by Swlabr
Can we not just apply the Burnside Basis Theorem and prove that all groups of order $p^2$ are Abelian? This would clearly prove the result.

Let $G$ be a finite $p$-group, and denote by $G'$ the derived subgroup, $G'=<[g,h]:g,h \in G>=$.

Denote by $\Phi(G)$ the Frattini subgroup; the intersection of all the maximal subgroups of $G$. Then $G/\Phi(G)$ is an elementary Abelian group which we can view as a vector space of rank $d(G)$. The Burnside Basis Theorem states:

i) $\Phi(G)=G'G^p$.

ii) For $S \subseteq G$ such that $=G$ then there exists $T \subseteq S$ such that $|T|=d(G)$ and $=G$.

iii) Another statement that isn't really relevant here but basically says that the generators of the group are precisely the coset representatives for the generators of the quotient group.

Thus, if $|\Phi(G)|=p$ we have that the group is cyclic (by part (ii) ), and if $|\Phi(G)|=1$ then we have that $G'=1$ and the group is Abelian (by part (i) ). Clearly there exists a subgroup of order $p$ in the group so the group cannot have $\Phi(G)=G$. Thus the group is Abelian.

I know it is a bit like overkill, but it is my favorite theorem...

The OP stated that the problem is given before the class equation, Sylow's theorems and stuff, so imagine Frattini's group!
Your point (ii) there though looks a little odd: it simply says that from a given set of generators of G, a finite group, you can always take a subset which equals the rank of the group=the minimal number of elements required to generate the group, which seems to be false (but I haven't given this too much thought), unless you meant that $T=\subset S\setminus \Phi(G)$, and anyway I don't see how from this it follows that if $|\Phi(G)|=p$ then G is cyclic...

Tonio

5. Originally Posted by tonio
The OP stated that the problem is given before the class equation, Sylow's theorems and stuff, so imagine Frattini's group!
Your point (ii) there though looks a little odd: it simply says that from a given set of generators of G, a finite group, you can always take a subset which equals the rank of the group=the minimal number of elements required to generate the group, which seems to be false (but I haven't given this too much thought), unless you meant that $T=\subset S\setminus \Phi(G)$, and anyway I don't see how from this it follows that if $|\Phi(G)|=p$ then G is cyclic...

Tonio
The group has to be a finite $p$-group. Essentially, the theorem says that if $G/\Phi(G)$ can be generated by $n$ elements then $G$ can be generated by $n$ elements. Thus, if $|G/\Phi(G)|=p$ then it is cyclic and so can be generated by $1$ element, which gives us that the group can be generated by $1$ element.

For example, look at $D_{16}$, the Dihedral group of order 16. We know that it is 2-generated. $\Phi(G)==G'$ which has order 4, and $G/G' \cong C_2 \times C_2$, a 2-generated group.

6. Originally Posted by Swlabr
The group has to be a finite $p$-group. Essentially, the theorem says that if $G/\Phi(G)$ can be generated by $n$ elements then $G$ can be generated by $n$ elements. Thus, if $|G/\Phi(G)|=p$ then it is cyclic and so can be generated by $1$ element, which gives us that the group can be generated by $1$ element.

Oh, this I do know: I've studied extensively the Frattini group of several tyes of groups, including pro-p groups.
What I didn't understand is your claim that "Thus, if we have that the group is cyclic (by part (ii) )"...how from part (ii)?

Tonio

7. Originally Posted by tonio
Oh, this I do know: I've studied extensively the Frattini group of several tyes of groups, including pro-p groups.
What I didn't understand is your claim that "Thus, if we have that the group is cyclic (by part (ii) )"...how from part (ii)?

Tonio
If $|\Phi(G)|=p$ then as $|G|=p^2$ we have that $|G/\Phi(G)|=p$. Thus, $G/\Phi(G)$ is cyclic and then we can apply the theorem to get that $G$ is cyclic.

8. Originally Posted by Swlabr
If $|\Phi(G)|=p$ then as $|G|=p^2$ we have that $|G/\Phi(G)|=p$. Thus, $G/\Phi(G)$ is cyclic and then we can apply the theorem to get that $G$ is cyclic.

I know this, and this is practically the same you wrote the last time, but you haven't yet answered my question: how does this follow from what you wrote in (ii) in your original message?!
This is my unique question here. In fact, I still don't understand exactly what your point (ii), and which isn't a part of Burnside's Basis Theorem as far as I am aware (but, of course, I could be wrong) actually means

Tonio

9. Originally Posted by tonio
I know this, and this is practically the same you wrote the last time, but you haven't yet answered my question: how does this follow from what you wrote in (ii) in your original message?!
This is my unique question here. In fact, I still don't understand exactly what your point (ii), and which isn't a part of Burnside's Basis Theorem as far as I am aware (but, of course, I could be wrong) actually means

Tonio
Yes, but I don't entirely understand your problem. What bit of it do you not understand? How you apply it to the problem, or the statement of it?

ii) For $S \subseteq G$ such that $=G$ then there exists $T \subseteq S$ such that $|T|=d(G)$ and $=G$.

For the problem, if $|G/\Phi(G)|=p$ then $G/\Phi(G)$ is cyclic. Thus, $d(G)=1$ and so if we take any subset $S \subseteq G$ such that $=G$ we can find another subset $T \subseteq S$ such that $|T|=d(G)=1$ and $=G$. Let [tex]T=\{g\}, so $G=$ is cyclic.

In Robinson, "A Course in the Theory of Groups", 5.3.2 has a much better statement of it than the one I gave:

Let $G$ be a finite $p$-group. Then $\Phi(G)=G'G^p$. Also, if $|G:\Phi(G)|=p^r$ every set of generators of $G$ has a subset of $r$ elements which also generates $G$.

Mine was taken, then modified through memory, from the one in Leedham-Green and McKay, "The Structure of Groups of Prime Power Order", Proposition 1.2.4.

10. Originally Posted by NonCommAlg
let $N=$ and $y \in G.$ if $o(y)=p^2,$ then $G$ would be cyclic, hence abelian, and we're done. also if $y=1_G,$ then $xy=yx.$ so we may assume that $o(y)=p.$ now we have $yxy^{-1}=x^j,$ for

some $j,$ because $N \lhd G.$ then $x=y^pxy^{-p}=x^{j^p}$ and therefore $x^{j^p - 1} = 1.$ hence $p \mid j^p -1,$ because $o(x)=p.$ but, by Fermat's little theorem, we also have that $p \mid j^p - j.$ thus $p \mid j - 1$ and

so $x^j = x.$ therefore $yxy^{-1}=x^j=x.$ hence $yx=xy. \ \Box$

Remark: a similar argument can be used to prove a more general result: if $G$ is any group of order $p^n,$ whre $p$ is prime, and $N \lhd G$ with $|N|=p,$ then $N \subseteq Z(G).$
Thanks NonCommAlg