Left regular representation of a group through group action

Question

Let $G$ be a group and let $g\cdot:G\to G$ (i.e., $g\cdot g'=gg'$). This induces a permutation representation of the group. I was trying to walk through the problems in Dummit and Foote. One of the problem statement was the following (pg 121, problem 3 of 4.2).

a) List the elements of $D_{8}$ and label them by $1,2,...,8$. Exhibit permutation representation of the group through action defined above. I knew this is a subgroup of $S_8$. Let's call this $A$.

b) Now relabel the element in a) as $1,3,5,7,2,4,6,8$ to get another permutation representation. This is isomorphic to $B\leq S_8$. Then they said $A$ is not isomorphic to $B$.

This problem is confusing to me as I do not see any reason why $A$ should not be isomorphic to $B$. My question is given two representations of a group acting upon the same set, should not two representations be completely isomorphic? Correct me if my understanding is wrong. I am doing self studying Dummit and Foote.

Answer 1

Yes, these two subgroups of $S_8$ are indeed isomorphic, though different (that is, not equal). Relabelling the points just gives you another "copy" of the group inside $S_8$.

However, this exercise is apparently being done with a good purpose in mind. My edition is slightly different (probably older) but, I think if you read ahead into the next section, you'll see the purpose of the exercise. More than just being isomorphic copies of the same (abstract) group, these two copies are what are called "conjugate". If you're not yet familiar with conjugacy, you can think of it as meaning that these two isomorphic copies of $D_8$ are isomorphic by a particular type of isomorphism connected with how they are situated inside $S_8$. In the edition I have, there is a follow-up exercise (problem 15 in 4.3) that asks you to determine exactly how they are related in this way.

Answer 2

Say $a$ and $b$ the two generators of $D_8$, of order $4$ and $2$ respectively. Then, Cayley's table reads: \begin{array}{l|cccccccc} & e & a & a^2 & a^3 & b & b a & b a^2 & b a^3 \\ \hline e & e & a & a^2 & a^3 & b & b a & b a^2 & b a^3 \\ a & a & a^2 & a^3 & e & b a^3 & b & b a & b a^2 \\ a^2 & a^2 & a^3 & e & a & b a^2 & b a^3 & b & b a \\ a^3 & a^3 & e & a & a^2 & b a & b a^2 & b a^3 & b \\ b & b & b a & b a^2 & b a^3 & e & a & a^2 & a^3 \\ b a & b a & b a^2 & b a^3 & b & a^3 & e & a & a^2 \\ b a^2 & b a^2 & b a^3 & b & b a & a^2 & a^3 & e & a \\ b a^3 & b a^3 & b & b a & b a^2 & a & a^2 & a^3 & e \end{array} When the authors say "List the elements of $D_8$ and label them by $1,2,\dots,8$", they mean to set the bijection $f\colon\{1,\dots,8\}\to D_8$ defined by: \begin{alignat}{1} &1&&\mapsto 1 \\ &2&&\mapsto a \\ &3&&\mapsto a^2 \\ &4&&\mapsto a^3 \\ &5&&\mapsto b \\ &6&&\mapsto ba \\ &7&&\mapsto ba^2 \\ &8&&\mapsto ba^3 \\ \tag1 \end{alignat} And when the authors say "Exhibit permutation representation of the group through action defined above.", they mean to calculate, for every $g\in D_8$, the permutation of $\{1,\dots,8\}$ defined by: $$f^{-1}l_gf \tag2$$ where $l_g$ is the left multiplication by $g$. For example, take $g=a$. Then, by $(1)$ and Cayley's table: \begin{alignat}{1} &(f^{-1}l_af)(1) &&= f^{-1}(l_a(f(1))) &&&= f^{-1}(a1) &&&&= f^{-1}(a) &&&&&=2 \\ &(f^{-1}l_af)(2) &&= f^{-1}(l_a(f(2))) &&&= f^{-1}(aa) &&&&= f^{-1}(a^2) &&&&&=3 \\ &(f^{-1}l_af)(3) &&= f^{-1}(l_a(f(3))) &&&= f^{-1}(aa^2) &&&&= f^{-1}(a^3) &&&&&=4 \\ &(f^{-1}l_af)(4) &&= f^{-1}(l_a(f(4))) &&&= f^{-1}(aa^3) &&&&= f^{-1}(1) &&&&&=1 \\ &(f^{-1}l_af)(5) &&= f^{-1}(l_a(f(5))) &&&= f^{-1}(ab) &&&&= f^{-1}(ba^3) &&&&&=8 \\ &(f^{-1}l_af)(6) &&= f^{-1}(l_a(f(6))) &&&= f^{-1}(aba) &&&&= f^{-1}(b) &&&&&=5 \\ &(f^{-1}l_af)(7) &&= f^{-1}(l_a(f(7))) &&&= f^{-1}(aba^2) &&&&= f^{-1}(ba) &&&&&=6 \\ &(f^{-1}l_af)(8) &&= f^{-1}(l_a(f(8))) &&&= f^{-1}(aba^3) &&&&= f^{-1}(ba^2) &&&&&=7 \\ \tag3 \end{alignat} Therefore, the element of $S_8$ which represents $a\in D_8$, with respect to the bijection $(1)$, is given by: $$(1234)(5876)$$ Etc. A different bijection than $(1)$ (your point b)) will give rise, for the same $g\in D_8$, to a different permutation of $\{1,\dots,8\}$ (though still of order $4$ for the element $a$). Though different, of course the two permutation groups are isomorphic, as both are isomorphic to $D_8$.