Interaction of the left and right regular representations of a group

Let \pi : G \rightarrow S_G be the left regular representation of G and let \lambda : G \rightarrow S_G be the right regular representation. That is, for all g, x \in G, \pi(g)(x) = gx and \lambda(g)(x) = xg^{-1}.

  1. Prove that \pi(g) and \lambda(h) commute for all g,h \in G. (Thus \pi[G] \leq C_{S_G}(\lambda[G]) and \lambda[G] \leq C_{S_G}(\pi[G]).)
  2. Prove that \pi(g) = \lambda(g) if and only if g \in G is an element of order 1 or 2 and g \in Z(G).
  3. Prove that \pi(g) = \lambda(h) if and only if g,h \in Z(G) and g = h^{-1}. Deduce that \pi[Z(G)] = \pi[G] \cap \lambda[G] = \lambda[Z(G)].

  1. Let g,h,x \in G. Now [\pi(g) \circ \lambda(h)](x) = \pi(g)(\lambda(h)(x)) = \pi(g)(xh^{-1}) = g(xh^{-1}) = (gx)h^{-1} = \lambda(h)(gx) = \lambda(h)(\pi(g)(x)) = [\lambda(h) \circ \pi(g)](x). Thus \pi(g) \circ \lambda(h) = \lambda(h) \circ \pi(g) for all g,h \in G. In other words, \pi[G] \leq C_{S_G}(\lambda[G]) and \lambda[G] \leq C_{S_G}(\pi[G]).
  2. (\Rightarrow) Suppose \pi(g) = \lambda(g). Then for all x \in G, gx = xg^{-1}. Letting x = 1 we see that g = g^{-1}; thus either g = 1 or |g| = 2. Moreover, we have gx = xg for all x \in G; thus g \in Z(G).

    (\Leftarrow) Suppose g \in Z(G) is either 1 or has order 2. Then g = g^{-1}. Then we have \pi(g)(x) = gx = xg = xg^{-1} = \lambda(g)(x) for all x \in G; hence \pi(g) = \lambda(g).

  3. (\Rightarrow) Suppose \pi(g) = \lambda(h). Then for all x \in G, gx = xh^{-1}. For x = 1 this gives g = h^{-1}. Thus we have gx = xg and hx = xh for all x \in G; hence g,h \in Z(G).

    (\Leftarrow) Suppose g,h \in Z(G) and g = h^{-1}. Then for all x \in G, we have \pi(g)(x) = gx = xg = xh^{-1} = \lambda(h)(x); thus \pi(g) = \lambda(h).

    Now if \sigma \in \pi[G] \cap \lambda[G], then \pi(g) = \sigma = \lambda(h) for some g, h \in G. As we saw above, g, h \in Z(G). Thus \pi[G] \cap \lambda[G] is contained in both \pi[Z(G)] and \lambda[Z(G)].

    Suppose \sigma \in \pi[Z(G)]. Then \sigma = \pi(g) for some g \in Z(G). Now \sigma(x) = \pi(g)(x) = gx = xg = \lambda(g^{-1})(x), hence \sigma \in \pi[G] \cap \lambda[G]. Similarly, \lambda[Z(G)] \subseteq \pi[G] \cap \lambda[G].

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Comments

  • Gobi Ree  On December 13, 2011 at 3:03 am

    In my book, g=h^{-1} is not in 3. Is it an error in the book?

    • nbloomf  On December 13, 2011 at 2:23 pm

      This is from the errata.

      I haven’t checked myself, but I imagine that the statement is false if we drop g=h^{-1}.

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