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This is Exercise 36 of section 4.3 (p. 132) from Abstract Algebra by Dummit & Foote.

Let $\DeclareMathOperator{\Sym}{Sym} \pi_1:G\rightarrow \Sym(G)$ be the left regular representation afforded by the action of $G$ on itself by left multiplication defined by $\pi_1(g)=\sigma_g$ so that $\sigma_g(x)=gx$ for all $x\in G$.

Let $\pi_2:G\rightarrow \Sym(G)$ be the right regular representation afforded by the action of $G$ on itself by right multiplication defined by $\pi_2(h)=\tau_h$ so that $\tau_h(x)=xh^{-1}$ for all $x\in G$.

Prove that $\sigma_g=\tau_h$ iff $g,h\in Z(G)$. Deduce that $\pi_1(G)\cap\pi_2(G)=\pi_1(Z(G))=\pi_2(Z(G))$

The 'only if ' part is as follows: Suppose $\sigma_g=\tau_h$. For all $x\in G$, $\sigma_g(x)=gx=xh^{-1}=\tau_h(x)$. In particular, $gh=hh^{-1}=1$, so $h=g^{-1}$. So $\sigma_g(x)=gx=xg=x(g^{-1})^{-1}=\tau_{g^{-1}}(x)$ for all $x\in G$. So $g\in Z(G)$. Also $h=g^{-1}\in Z(G)$.

As you can see, the 'only if' part is quite easy. What's frustrating me is the 'if' part. Whenever $g$ and $h$ lie in the center of $G$, why is $\sigma_g=\tau_h$? This is so counter-intuitive: Since $1\in Z(G)$, by letting $g\in Z(G)$, $g\neq 1$, we get $gx=x$ for all $x\in G$, and so $g=1$! And so $G$ is centerless!

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  • $\begingroup$ The "if part" is false. If g and g are central, then left multiplication by g is the same as right multiplication by the inverse of g, and the latter is the same as right multiplication by the inverse of h iff g ans $h^{-1}$ are equal $\endgroup$ – Mariano Suárez-Álvarez Jun 17 '17 at 4:02
  • $\begingroup$ So you mean there is a typo in the book? It's not uncommon though. I've seen 'if and only if' being typed as 'if and only'. $\endgroup$ – user441558 Jun 17 '17 at 5:17
  • $\begingroup$ there is no typo in the book , in atleast my D&F copy. :"iff g,h $\in Z(G)$ and $g=h^{-1}$".This is the question $\endgroup$ – Riju Jun 17 '17 at 6:32
  • $\begingroup$ @Riju There's no such '$g=h^{-1}$' in my copy. $\endgroup$ – user441558 Jun 17 '17 at 6:59
  • $\begingroup$ As you've already seen (indicated in your last paragraph), the disproof of "$\sigma_g=\tau_h$ if $g,h\in Z(G)$" is quite easy, so indeed it is a typo. Thus everything appears cleared up now. $\endgroup$ – arctic tern Jun 18 '17 at 3:57

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