# grop of order p^3 , group homomorphism

• Nov 23rd 2010, 01:19 PM
hermanni
grop of order p^3 , group homomorphism
Prove that if p is an odd prime and P is a group of order $p^{3}$ then the $p^{th}$ power map $x -> x^{p}$ is a homomorphism of P into Z(P). If P is not cyclic , show that the kernel of the $p^{th}$ map has order $p^{2}$ or $p^{3}$. Is the squaring map a homomorphism in nonabelian groups of order 8? Where the oddness of p needed in the above proof?
Hint: Assume x, y in G and both x and y commute with $[x,y]=x^{-1} y^{-1} xy$. Prove that for all n in Z+ , $(xy)^{n} = x^{n} y^{n} [y,x]^{(n(n-1)/2)}$

My problem is I can't see why the hint holds.I aldo don't understand where we
• Nov 23rd 2010, 01:34 PM
tonio
Quote:

Originally Posted by hermanni
Prove that if p is an odd prime and P is a group of order $p^{3}$ then the $p^{th}$ power map $x -> x^{p}$ is a homomorphism of P into Z(P). If P is not cyclic , show that the kernel of the $p^{th}$ map has order $p^{2}$ or $p^{3}$. Is the squaring map a homomorphism in nonabelian groups of order 8? Where the oddness of p needed in the above proof?
Hint: Assume x, y in G and both x and y commute with $[x,y]=x^{-1} y^{-1} xy$. Prove that for all n in Z+ , $(xy)^{n} = x^{n} y^{n} [y,x]^{(n(n-1)/2)}$

My problem is I can't see why the hint holds.I aldo don't understand where we

The hint in itself is a nice exercise, try it by induction. Now, a non-abelian group of order $p^3$ (if the group is abelian

the exercise is trivial) has center of order p, which is exactly the same as the derived subgroup of G.

So $\left|G/Z(G)\right|=p\Longrightarrow x^p\in Z(G)\,,\,\forall x\in G$ . try to take it from here.

Tonio
• Nov 23rd 2010, 01:42 PM
topspin1617
Quote:

Originally Posted by tonio
The hint in itself is a nice exercise, try it by induction. Now, a non-abelian group of order $p^3$ (if the group is abelian

the exercise is trivial) has center of order p, which is exactly the same as the derived subgroup of G.

So $\left|G/Z(G)\right|=p\Longrightarrow x^p\in Z(G)\,,\,\forall x\in G$ . try to take it from here.

Tonio

Wait... if $\left|Z(G)\right|=p$, then shouldn't $\left|G/Z(G)\right|=p^3/p=p^2$?

Even then, of course, this would force $G/Z(G)$ to be abelian. There are only 2 (abelian) groups of order $p^2$; cyclic and elementary abelian. The group $G/Z(G)$ cannot be cyclic, though, else $G$ would be abelian. Thus

$G/Z(G)\cong \mathbb{Z}_p\times \mathbb{Z}_p\Longrightarrow x^p\in Z(G)\,\,\forall\,\, x\in G$,

so we get what you wanted to conclude from that fact anyway.
• Nov 23rd 2010, 01:59 PM
tonio
Quote:

Originally Posted by topspin1617
Wait... if $\left|Z(G)\right|=p$, then shouldn't $\left|G/Z(G)\right|=p^3/p=p^2$?

Even then, of course, this would force $G/Z(G)$ to be abelian. There are only 2 (abelian) groups of order $p^2$; cyclic and elementary abelian. The group $G/Z(G)$ cannot be cyclic, though, else $G$ would be abelian. Thus

$G/Z(G)\cong \mathbb{Z}_p\times \mathbb{Z}_p\Longrightarrow x^p\in Z(G)\,\,\forall\,\, x\in G$,

so we get what you wanted to conclude from that fact anyway.

Of course, my mistake. Thanx.

Tonio
• Nov 23rd 2010, 02:12 PM
hermanni
Thanx , both of you guys (Rofl)
• Nov 23rd 2010, 02:27 PM
topspin1617
I'm gonna give more of a hint, because I thought about this one myself and I think the way to proceed was not at all obvious (unless there's an easier way I missed).

Let's assume the hint. Note, then, that to prove that the $p^{th}$-power map is a homomorphism, it suffices (by the hint) to prove that $[y,x]^{(p(p-1)/2)}=e$ (the identity of $P$) for all $x,y\in P$.

Assume that $P$ is nonabelian (otherwise the result of the problem is trivial). First, we find the order of $Z(P)$. All $p$-groups have nontrivial centers, and $P$ is nonabelian; thus, $|Z(P)|\notin \{ 1,p^3\}$. If $|Z(P)|=p^2$, then $|P/Z(P)|=p\Rightarrow P/Z(P)$ is cyclic $\Rightarrow P$ is abelian, obviously a contradiction.

So we conclude $|Z(P)|=p$.

Then $|P/Z(P)|=p^2\Rightarrow P/Z(P)$ is abelian.

Recall the following properties of the commutator subgroup $G^{\prime}$ of a group $G$:
• If $N\trianglelefteq G$ such that the quotient group $G/N$ is abelian, then $G^{\prime}\leq N$.
• $G^{\prime}=\{e\}$ if and only if $G$ is abelian.

Using this, can we say what the order of $P^{\prime}$ is? And what does this say about the equation in the hint, when $n=p$?
• Nov 24th 2010, 12:31 AM
Swlabr
I believe we can use the Burnside Basis Theorem here (see p140 of Robinson's book, A Course in the Theory of Groups'). It actually yields quite a neat proof, which is why I post this...it is also one of my favourite theorems!

Now, because we are in a non-abelian group of order $p^3$ we must have that $|\Phi(G)|=p$, by the BBT.

Also by the BBT we have that $\Phi(G) = G^{\prime}G^p$.

Therefore, it is sufficient to prove that $Z(G) \leq G^{\prime}$, which it is (see topspin's post).

I am, however, a bit confused as to where we need odd primes. Certainly this proof does not rely on odd primes, glancing over at p141 of Robinson I see that the identity in the hint doesn't need odd primes either, and I'm pretty sure topspin's proof doesn't need odd primes...also, the result does hold for the Quaternion group, and for $D_8$, the two non-abelian groups of order 8...

EDIT: for notation, $\Phi(G) = Frat(G)=$ the Frattini subgroup $= \bigcap_{M\text{ maximal in }G} M$.
• Nov 24th 2010, 03:36 AM
tonio
Quote:

Originally Posted by Swlabr
I believe we can use the Burnside Basis Theorem here (see p140 of Robinson's book, A Course in the Theory of Groups'). It actually yields quite a neat proof, which is why I post this...it is also one of my favourite theorems!

Now, because we are in a non-abelian group of order $p^3$ we must have that $|\Phi(G)|=p$, by the BBT.

Also by the BBT we have that $\Phi(G) = G^{\prime}G^p$.

Therefore, it is sufficient to prove that $Z(G) \leq G^{\prime}$, which it is (see topspin's post).

I am, however, a bit confused as to where we need odd primes. Certainly this proof does not rely on odd primes, glancing over at p141 of Robinson I see that the identity in the hint doesn't need odd primes either, and I'm pretty sure topspin's proof doesn't need odd primes...also, the result does hold for the Quaternion group, and for $D_8$, the two non-abelian groups of order 8...

EDIT: for notation, $\Phi(G) = Frat(G)=$ the Frattini subgroup $= \bigcap_{M\text{ maximal in }G} M$.

If you allow the prime $p=2$ in this problem, then you get that $x\rightarrow x^2$ is a homomorphism, which

gives us $xyxy=(xy)^2=x^2y^2\Longrightarrow yx=xy\Longrightarrow$ the group is abelian.

Tonio
• Nov 24th 2010, 04:14 AM
Swlabr
Quote:

Originally Posted by tonio
If you allow the prime $p=2$ in this problem, then you get that $x\rightarrow x^2$ is a homomorphism, which

gives us $xyxy=(xy)^2=x^2y^2\Longrightarrow yx=xy\Longrightarrow$ the group is abelian.

Tonio

Ah, yes, of course. I was just seeing if squaring took you into the center for p=2, not to see if it was a homomorphism.