Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Map is given, for any element $x \mapsto x^p$. I proved that the image of this map is contained in center of $G$. It is of course true when $G$ is commutative. So assume that is is not that case, $G/C(G)$ must be isomorphic to $Z/pZ\times Z/pZ$. Let the generators $aC(G)$, $bC(G)$. $C(G)=\langle c\rangle$. By using the definition of $C(G)$, for $x \in G$, write an combinations of $a,b,c$ and it is easy to show that $x^pC(G)=C(G)$ this is what i found. but still, It does not give an answer to my first question i.e. i want to show that $x^py^p=(xy)^p$

share|improve this question
1  
When p=2 things go wrong. In general when G is finite, $x \mapsto x^n$ is a homomorphism from G to its center for $n=[G:Z(G)]$. This is $n=p^2$ though, and so is a pretty trivial map in this case. –  Jack Schmidt Sep 17 '12 at 21:58

2 Answers 2

You know that if $G$ is a group and $x,y\in G$ be two arbitrary elements of it, then $$[x,y]=x^{-1}y^{-1}xy$$ is called a commutator of $G$. All such these elements in $G$, when $x,y\in G$ construct a well-known subgroup called the commutator subgroup $G'$. Now, there are two lemmas which can be proved by induction on $n$:

Lemma 1. Let $G$ is a group and let the element $[x,y^{-1}]$ can be commute with $x$ and $y$, then: $$[x,y^{-n}]=[x,y^{-1}]^n$$

and by using it we have the second:

Lemma 2. Let $G$ is a group and let the element $[x,y^{-1}]$ can be commute with $x$ and $y$, then: $$(xy)^n=x^ny^n[x,y^{-1}]^{\frac{n(n-1)}{2}}$$

What we have is:

$|G|=p^3$ leads $Z(G)=G'$ and $|Z(G)|=|G'|=p$ since $G$ is non- abelian finite group.

$Z(G)=G'$ means the element $[x,y^{-1}]$ is central and then by lemma 2 you will have: $$(xy)^p=x^py^p[x,y^{-1}]^{\frac{p(p-1)}{2}}=x^py^p\big([x,y^{-1}]^p\big)^{\frac{(p-1)} {2}}$$ $p$ is an odd prime number so $\frac{(p-1)} {2}$ is an integer and from $|G'|=p$ we have $$[x,y^{-1}]^p=1$$ This means to us that $$(xy)^p=x^py^p$$ This is what you have been searching for.

share|improve this answer

Let $x,y \in G$. As $G/C(G)$ is commutative, $xyC(G) = yxC(G)$, so $yx = xyz$ for some $z \in C(G)$. We have by induction $y^kx = xy^kz^k$ as: \[ y^kx = y^{k-1}yx = y^{k-1}xyz = xy^kz^{k-1}z = xy^kz^k \] No we prove by induction that $(xy)^k = x^ky^kz^{\frac 12k(k-1)}$, which holds since \[ (xy)^k = (xy)^{k-1}xy = x^{k-1}y^{k-1}xyz^{\frac 12(k-1)(k-2)} = x^ky^kz^{\frac 12(k-1)(k-2) + (k-1)} = x^ky^kz^{\frac 12k(k-1)} \] So $(xy)^p = x^py^pz^{\frac 12p(p-1)}$. If $p$ is odd, then the exponent is a multiple of $p$ (which is the order of $C(G)$), hence $z^{\frac 12 p(p-1)} = 1$,and so $(xy)^p = x^py^p$.

If $p$ is even, this needn't hold true: In the quaternion group of order $8 = 2^3$, we have $(ij)^2 = k^2 = -1 \ne 1 = (-1)^2 = i^2j^2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.