**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.4**

For each of the groups $S_3$, $D_8$, and $Q_8$, compute the centralizer of each element and find the center of the group.

Solution: We prove some lemmas which will cut down on some of the work we have to do.

**Lemma 1**: Let $G$ be a group and $A \subseteq G$. Then $g \in C_G(A)$ if and only if $g^{-1} \in C_G(A)$.

Proof: Since $C_G(A)$ is a group, it is closed under inversion. $\square$

**Lemma 2**: Let $G$ be a group and $x,y \in G$. Then $y \in C_G(x)$ if and only if $x \in C_G(y)$.

Proof: We have $xyx^{-1} = y$ if and only if $xy = yx$ if and only if $x = yxy^{-1}$. $\square$

**Lemma 3**: Let $G$ be a group and $x \in G$. Then $C_G(x) = C_G(x^{-1})$.

Proof: If $y \in C_G(x)$, then we have $yxy^{-1} = x$, so that $x = y^{-1}xy$. Inverting both sides gives $x^{-1} = yx^{-1}y^{-1}$. The other inclusion is similar. $\square$

**Lemma 4**: Let $G$ be a group. Then $Z(G) = \bigcap_{x \in G} C_G(x)$.

Proof: ($\subseteq$) If $y \in Z(G)$, then $yx = xy$ for all $x \in G$, hence $x = yxy^{-1}$ for all $x \in G$. So $y \in C_G(x)$ for all $x \in G$, thus $y \in \bigcap_{x \in G} C_G(x)$. ($\supseteq$) If $y \in \bigcap_{x \in G} C_G(x)$, then we have $x = yxy^{-1}$ for all $x \in G$, so that $xy = yx$ for all $x \in G$. Thus $y \in Z(G)$. $\square$

**Lemma 5**: Let $n \in \mathbb{Z}^+$, and consider $D_{2n}$. We have $sr^a \in C_G(r^b)$ if and only if $b = -b \pmod n$.

Proof: We know from a previous exercise that $(sr^a)^{-1} = sr^a$. Now $sr^ar^bsr^a = sr^{-b}$, and since elements of $D_{2n}$ have unique representations, $r^b = r^{-b}$ if and only if $b = -b \pmod n$. $\square$

**Lemma 6**: Let $n \in \mathbb{Z}^+$, and consider $D_{2n}$. We have $sr^a \in C_G(sr^b)$ if and only if $2a = 2b \pmod n$.

Proof: We have $sr^asr^bsr^a = sr^{2a-b}$. Since elements of $D_{2n}$ have unique representations (mod n) we have $2a = 2b \pmod n$. $\square$

$S_3$

$x$ | Reasoning | ------------$C_G(x)$------------ |

1 | $S_3$ | |

(1 2) | (1 2 3)(1 2)(1 3 2) = (2 3) (1 3)(1 2)(1 3) = (2 3) (2 3)(1 2)(2 3) = (1 3) | $\{1, (1\ 2)\}$ |

(1 3) | (1 2 3)(1 3)(1 3 2) = (1 2) (2 3)(1 3)(2 3) = (1 2) | $\{ 1, (1\ 3) \}$ |

(2 3) | (1 2 3)(2 3)(1 3 2) = (1 3) | $\{ 1, (2\ 3) \}$ |

(1 2 3) | Lemma 2 | $\{ 1, (1\ 2\ 3), (1\ 3\ 2) \}$ |

(1 3 2) | Lemma 3 | $\{ 1, (1\ 2\ 3), (1\ 3\ 2) \}$ |

$D_8$

$x$ | Reasoning | ------$C_G(x)$------ |

1 | $D_8$ | |

$r$ | Since $-1 = 3 \pmod 4$, $sr^a \notin C_G(r)$ by Lemma 5. | $\{ 1, r, r^2, r^3 \}$ |

$r^2$ | Exercise 2.2.3 | $D_8$ |

$r^3$ | Lemma 3 | $\{ 1, r, r^2, r^3 \}$ |

$s$ | Lemma 2 and Lemma 6 since $2a = 0 \pmod 4$ has solutions 0 and 2. | $\{ 1, s, r^2, sr^2 \}$ |

$sr$ | Lemma 2 and Lemma 6 since $2a = 2 \pmod 4$ has solutions 1 and 3. | $\{ 1, sr, r^2, sr^3 \}$ |

$sr^2$ | Lemma 2 and Lemma 6 since $2a = 0 \pmod 4$ has solutions 0 and 2. | $\{ 1, s, r^2, sr^2 \}$ |

$sr^3$ | Lemma 2 and Lemma 6 since $2a = 2 \pmod 4$ has solutions 1 and 3. | $\{ 1, sr, r^2, sr^3 \}$ |

$Q_8$

$x$ | Reasoning | ------------$C_G(x)$------------ |

1 | $Q_8$ | |

$i$ | $ji(-j) = -i$, $ki(-k) = -i$ | $\{ 1, i, -1, -i \}$ |

$-i$ | Lemma 3 | $\{ 1, i, -1, -i \}$ |

$j$ | Lemma 2 and $kj(-k) = -j$ | $\{ 1, j, -1, -j \}$ |

$-j$ | Lemma 3 | $\{ 1, j, -1, -j \}$ |

$k$ | Lemma 2 | $\{ 1, k, -1, -k \}$ |

$-k$ | Lemma 3 | $\{ 1, k, -1, -k \}$ |

$-1$ | Lemma 2 | $Q_8$ |

Now using Lemma 4, we can see by inspection that $Z(S_6) = 1$, $Z(D_8) = \{ 1, r^2 \}$, and $Z(Q_8) = \{ 1, -1 \}$.