Commutators and Groups of Prime-Cubed Order (5.4.7-9)

Dummit and Foote Abstract Algebra, section 5.4, exercises 7-9:

7. Let $p$ be prime and let $P$ be a non-abelian group of order $p^3$. Prove $P'=Z(P)$.

8. Let $x,y∈G$ commute with $[x,y]$. Prove that for all $n∈\mathbb{Z}^+$, we have $(xy)^n = x^ny^n[y,x]^{n(n-1)/2}$.

9. Let $p$ be an odd prime, $φ$ be the $p$-power map $φ : x \mapsto x^p$, and $P$ be a group of order $p^3$. Prove $φ$ is a homomorphism and $\text{im}~φ ≤ Z(P)$. If $P$ is not cyclic, show that $|\text{ker}~φ|∈\{p^2,p^3\}$. Is $φ$ a homomorphism in non-abelian groups of order 8? Where is the oddness of $p$ invoked in the above proof? [Use exercise 7.]

Proof: (7) By the class equation we have $|Z(P)|≠1$ and by the non-abelianness of $P$ we have $|Z(P)|≠p^3$. Assuming $|Z(P)|=p^2$ leads to $P/Z(P)$ being cyclic and thus $P$ being abelian, so therefore $|Z(P)|=p$. By previous investigations and/or similar logic, all groups of order $p^2$ are abelian, so that $P/Z(P)$ is abelian and thus $P'≤Z(P)$. This entails either $P'$ is trivial or $P'=Z(P)$; the former implies $P/1≅P$ is abelian, so the latter is the only available possibility.

(8) First note that if $x[x,y]x^{-1}=[x,y]$, then $(x[x,y]x^{-1})^{-1}=[x,y]^{-1}$, so that by exercise 1 the right-hand side is $[y,x]$ and the left-hand side is $(x[x,y]x^{-1})^{-1}=x[x,y]^{-1}x^{-1}=x[y,x]x^{-1}$, so that $x,y$ in this exercise might be assumed to commute with $[y,x]$ as well.

Proceed by induction. The basis case $(xy)^1=xy=x^1y^1[y,x]^{(1)(0)/2}$ holds true, so assume $(xy)^k=x^ky^k[y,x]^{k(k-1)/2}$ for all $k < n$. We have $(xy)^n=(xy)^{n-1}xy=x^{n-1}y^{n-1}[y,x]^{(n-1)(n-2)/2}xy=x^{n-1}y^{n-1}xy[y,x]^{(n-1)(n-2)/2}$$=x^{n-1}y^{n-1}xy[x,y]^{n-1}[y,x]^{n-1}[y,x]^{(n-1)(n-2)/2}=x^{n-1}y^{n-1}xy[x,y]^{n-1}[y,x]^{n(n-1)/2}$. This implies that the truth of the original proposition hinges on the validity of the new proposition, $x^{n-1}y^{n-1}xy[x,y]^{n-1}=x^ny^n$. Rework this to $xy^n=y^{n-1}xy[x,y]^{n-1}$, and proceed with further induction. The basis case with $n=1$ clearly holds, so apply the inductive hypothesis and finish the problem: $xy^n=xy^{n-1}y=y^{n-2}xy[x,y]^{n-2}y$$=y^{n-2}xy[y,x][x,y]^{n-1}y=y^{n-2}xy[y,x]y[x,y]^{n-1}=y^{n-1}xy[x,y]^{n-1}$.

(9) Assume $P$ is abelian. Then $φ(xy) = (xy)^p = x^py^p = φ(x)φ(y)$ is a homomorphism and since $Z(P) = P$ we clearly have its image contained therein. Therefore assume $P$ is non-abelian. Since by exercise 7, we have $P'=Z(P)$, the setting verifies the criterion for exercise 8 that any $x,y∈P$ commute with $[x,y]∈P'$. Since $p$ is odd, we have $2 \mid p-1$, so that $p \mid p(p-1)/2$. This implies $[y,x]^{p(p-1)/2}=1$ since $P'$ is a group of order $p$, and thus we have the equation from the previous exercise collapsing to $(xy)^p=x^py^p$, and $φ$ is thus a homomorphism.

Assume $φ(x) \not ∈ Z(P)$ for some $x∈P$. This implies $\bar{x}^p≠1$ in $P/Z(P)$, and since $|P/Z(P)|=p^2$, that $\bar{x}$ generates this quotient group, so that it is cyclic, and $P$ is abelian by the familiar process. Therefore, $\text{im}~φ ≤ Z(P)$.

By Cauchy's theorem we have $x$ of order $p$ so that the kernel is not the identity. If the kernel is of order $p$, then the image is of order $p^2$. Since the image being cyclic would imply $|y^p|=p^2$ so $|y|=p^3$ for some $y$, we have the image and thus $P$ contains $p^2-1$ elements of order $p$, and thus the kernel's order is greater than $p$.

As for the oddness of $p$, it was invoked in collapsing the formula in exercise 8. It does indeed map the elements of the only two non-abelian groups of order 8 ($D_8$ and $Q_8$) to their respective centers, but it is not a homomorphism on either, as $φ(s)=1$ and $φ(r)=r^2$, but $φ(rs)=1≠r^2$, and $φ(i)=φ(j)=-1$, but $φ(ij)=-1≠1$. $\square$

The proof to exercise 8 might not be the shortest, though it was not a random long shot. The first induction simplified the problem by replicating the tail of what it ultimately was aiming for and thus appealed to a simpler inductive problem, in a way exhibiting an inductive process itself.