Find the centralizer of each element in Dih(8), the quaternion group, Sym(3), and Dih(16)

Use the given subgroup lattices to find the centralizers of every element of the following groups.

  1. D_8
  2. Q_8
  3. S_3
  4. D_{16}

Lemma 1: Let G be a group, let x \in G, and let z \in Z(G). Then C_G(xz) = C_G(x). Proof: Suppose y \in C_G(x); then yxz = xyz = xzy, so y \in C_G(xz). If y \in C_G(xz), we have yxz = xzy = xyz, so that by cancellation, yx = xy and we have y \in C_G(x). \square

Lemma 2: Let G be a group and x,g \in G. Then C_G(gxg^{-1}) = g(C_G(x))g^{-1}. Proof: Suppose y \in C_G(x). Then y = gzg^{-1} for some z \in C_G(x). Then (gzg^{-1})(gxg^{-1}) = gzxg^{-1} = gxzg^{-1} = (gxg^{-1})(gzg^{-1}), so that y \in C_G(gxg^{-1}). Suppose y \in C_G(gxg^{-1}). Then ygxg^{-1} = gxg^{-1}y, so that (g^{-1}yg)x = x(g^{-1}yg). Thus g^{-1}yg \in C_G(x), and we have y \in g C_G(x) g^{-1}. \square

Lemma 3: Let G be a group and let g \in G, A \subseteq G. Then g \langle A \rangle g^{-1} = \langle g A g^{-1} \rangle. Proof: Let y \in g \langle A \rangle g^{-1}. Then y = gzg^{-1} where z \in \langle A \rangle; recall that we may write z = a_1 a_2 \ldots a_n where for each i, either a_i or a_i^{-1} is in A. Then z = a_1g^{-1}ga_2g^{-1}g\ldots g^{-1}ga_n, so that gzg^{-1} = (ga_1g^{-1})(ga_2g^{-1}\ldots(ga_ng^{-1}). Hence y \in \langle gAg^{-1} \rangle. All of these steps are “if and only if”, so the sets are equal. \square

Lemma 4: Let G be a group and let a,b \in G. If \langle a \rangle = \langle b \rangle, then C_G(a) = C_G(b). Proof: Note that b = a^k for some k, so that if xa = ax, we have xb = bx. Hence C_G(a) \leq C_G(b). The other direction is similar. \square

  1. G = D_8. Recall that Z(D_8) = \{ 1, r^2 \}.
    x Reasoning C_G(x)
    1 1 \in Z(D_8) D_8
    r \langle r \rangle \leq C_G(r), so C_G(r) is either \langle r \rangle or D_8. But sr \neq rs, so s \notin C_G(r), hence C_G(r) \neq D_8. \langle r \rangle
    r^2 r^2 \in Z(D_8) D_8
    r^3 r^3 = r^{-1} \langle r \rangle
    s \langle s \rangle \leq C_G(s) and \langle r^2 \rangle \leq C_G(s) since r^2 \in Z(G), so C_G(s) is either \langle s,r^2 \rangle or D_8. But r \notin C_G(s) since sr \neq rs. \langle s,r^2 \rangle
    sr \langle sr \rangle \leq C_G(sr) and \langle r^2 \rangle \leq C_G(sr) since r^2 \in Z(G), so C_G(sr) is either \langle sr, r^2 \rangle or D_8. But rsr = s \neq sr^2, so r \notin C_G(sr). \langle sr, r^2 \rangle
    sr^2 sr^2 = s \cdot r^2, so C_G(sr^2) = C_G(s) by Lemma 1 above. \langle s,r^2 \rangle
    sr^3 sr^3 = sr \cdot r^2, so C_G(sr^3) = C_G(sr) by Lemma 1 above. \langle sr,r^2 \rangle
  2. G = Q_8. Recall that Z(Q_8) = \{ 1, -1 \}.
    x Reasoning C_G(x)
    1 1 \in Z(G) Q_8
    -1 -1 \in Z(G) Q_8
    i \langle i \rangle \leq C_G(i) so C_G(i) is either \langle i \rangle or Q_8. But j \notin C_G(i) since ij = k \neq -k = ji. \langle i \rangle
    -i -i = i^{-1}, so C_G(-i) = C_G(i) \langle i \rangle
    j \langle j \rangle \leq C_G(j) so C_G(j) is either \langle j \rangle or Q_8. But k \notin C_G(j) since jk = i \neq -i = kj. \langle j \rangle
    -j -j = j^{-1}, so C_G(-j) = C_G(j). \langle j \rangle
    k \langle k \rangle \leq C_G(k), so C_G(k) is either \langle k \rangle or Q_8. But i \notin C_G(k) since ki = j \neq -j = ik. \langle k \rangle
    -k -k = k^{-1}, so C_G(-k) = C_G(k). \langle k \rangle
  3. S_3
    x C_G(x)
    1 S_6
    (1\ 2) \langle (1\ 2) \rangle \leq C_G((1\ 2)), so C_G((1\ 2)) is either S_6 or \langle (1\ 2) \rangle. By a lemma to a previous exercise, there exists an element of S_6 which does not commute with (1\ 2). \langle (1\ 2) \rangle
    (1\ 3) Note that (1\ 3) = (2\ 3)(1\ 2)(2\ 3), so we can apply Lemmas 2 and 3. \langle (1\ 3) \rangle
    (2\ 3) Note that (2\ 3) = (1\ 3)(1\ 2)(1\ 3), so we can apply Lemmas 2 and 3. \langle (2\ 3) \rangle
    (1\ 2\ 3) \langle (1\ 2\ 3) \leq C_G((1\ 2\ 3)), so C_G((1\ 2\ 3)) is either \langle (1\ 2\ 3) \rangle or S_6. But (1\ 2) does not commute with (1\ 2\ 3). \langle (1\ 2\ 3) \rangle
    (1\ 3\ 2) (1\ 3\ 2) = (1\ 2\ 3)^{-1} \langle (1\ 2\ 3) \rangle
  4. D_{16}. Recall that, as we saw in a previous exercise, r^4 commutes with all elements of D_{16}.
    x C_G(x)
    1 D_{16}
    r \langle r \rangle \leq D_{16}, so C_G(r) is either D_{16} or \langle r \rangle. But rs \neq sr. \langle r \rangle
    r^2 \langle r^2 \rangle \leq C_G(r^2), so C_G(r) is either \langle r^2 \rangle, \langle s,r^2 \rangle, \langle r \rangle, \langle sr,r^2 \rangle, or D_{16}. Note that sr^2 \neq r^2 s, and rr^2 = r^2r, and srr^2 \neq r^2sr. \langle r \rangle
    r^3 \langle r^3 \rangle = \langle r \rangle, so Lemma 4 applies. \langle r \rangle
    r^4 r^4 \in Z(G). D_{16}
    r^5 \langle r^5 \rangle = \langle r \rangle, so Lemma 4 applies. \langle r \rangle
    r^6 r^6 = r^2 r^4 \langle r \rangle
    r^7 r^7 = r^{-1}. \langle r \rangle
    s \langle s \rangle \leq C_G(s), so C_G(s) is either \langle s \rangle, \langle s, r^4 \rangle, \langle s,r^2 \rangle, or D_{16}. Now \langle r^4 \rangle \leq C_G(s), and r^2s \neq sr^2. \langle s,r^4 \rangle
    sr \langle sr \rangle \leq C_G(sr), so C_G(sr) is either \langle sr \rangle, \langle sr,r^4 \rangle, \langle sr,r^2 \rangle, or D_{16}. Note that r^4 sr = srr^4 and srr^2 \neq r^2sr. \langle sr,r^4 \rangle
    sr^2 \langle sr^2 \rangle \leq C_G(sr^2), so C_G(sr^2) is either \langle sr^2 \rangle, \langle sr^2,r^4 \rangle, \langle s,r^2 \rangle, or D_{16}. Note that r^4sr^2 = sr^2r^4 and r^2sr^2 \neq sr^2r^2. \langle sr^2,r^4 \rangle
    sr^3 \langle sr^3 \rangle \leq C_G(sr^3) and \langle r^4 \rangle \leq C_G(sr^3). Note that r^2 sr^3 \neq sr^3r^2. \langle sr^3,r^4 \rangle
    sr^4 Lemma 1 \langle s,r^4 \rangle
    sr^5 Lemma 1 \langle sr,r^4 \rangle
    sr^6 Lemma 1 \langle sr^2,r^4 \rangle
    sr^7 Lemma 1 \langle sr^3,r^4 \rangle
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  • Bobby Brown  On September 19, 2010 at 8:07 pm

    typo in centralizer of sr^3.

    • nbloomf  On September 19, 2010 at 9:04 pm

      Fixed. Thanks!

  • Gobi Ree  On November 17, 2011 at 1:34 am

    It seems that the typo in centralizer of sr^3 still exists.

    • nbloomf  On November 17, 2011 at 10:55 am


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