Compute the normalizer of each subgroup in Sym(3) and the quaternion group

In each of the following groups find the normalizer of each subgroup.

(1) S_3 (2) Q_8


Lemma 1: For all groups G, N_G(1) = G. Proof: If g \in G, we have g \cdot 1 \cdot g^{-1} = 1, so G \subseteq N_G(1). \square

Lemma 2: For all groups G, N_G(G) = G. Proof: Clearly if g \in G we have gGg^{-1} = G. \square

Lemma 3: Let G be a group and A \subseteq Z(G). Then N_G(A) = G. Proof: Let g \in G and a \in A. Then gag^{-1} = a, so that gAg^{-1} = A. Hence G \subseteq N_G(A). \square

Lemma 4: Let G be a group, h \in G, and A \subseteq G. Then N_G(hAh^{-1}) = h N_G(A) h^{-1}. Proof: We have g \in N_G(hAh^{-1}) iff ghAh^{-1}g^{-1} = hAh^{-1}, iff (h^{-1}gh)A(h^{-1}gh) = A, iff h^{-1}gh \in N_G(A), iff g \in h N_G(A) h^{-1}. \square

Lemma 5: If H \leq G is a subgroup, then H \leq N_G(H). Proof: Clearly if h \in H, hHh^{-1} = H. \square

Lemma 6: Let G be a group, a \in G, and g \in G. If \langle gag^{-1} \rangle = \langle a \rangle, then g \in N_G(\langle a \rangle). Proof: We saw in a lemma to a previous exercise that g \langle a \rangle g^{-1} = \langle gag^{-1} \rangle = \langle a \rangle, so that g \in N_G(\langle a \rangle). \square

  1. G = S_6
    H Reasoning N_G(H)
    1 Lemma 1 S_3
    \langle (1\ 2) \rangle \langle (1\ 2) \rangle \leq N_G(\langle (1\ 2) \rangle), so N_G(\langle (1\ 2) \rangle) is either \langle (1\ 2) \rangle or S_3. But (1\ 2\ 3)(1\ 2)(1\ 3\ 2) = (2\ 3). \langle (1\ 2) \rangle
    \langle (1\ 3) \rangle (1\ 3\ 2)(1\ 2)(1\ 2\ 3) = (1\ 3), so Lemma 4 applies. \langle (1\ 3) \rangle
    \langle (2\ 3) \rangle (1\ 2\ 3)(1\ 2)(1\ 3\ 2) = (2\ 3), so Lemma 4 applies. \langle (2\ 3) \rangle
    \langle (1\ 2\ 3) \rangle \langle (1\ 2\ 3) \rangle \leq N_G(\langle (1\ 2\ 3) \rangle), so N_G(\langle (1\ 2\ 3) \rangle) is either \langle (1\ 2\ 3) \rangle or S_3. Note that (1\ 2)(1\ 2\ 3)(1\ 2) = (1\ 3\ 2) and (1\ 2)(1\ 3\ 2)(1\ 2) = (1\ 2\ 3), so that (1\ 2) \in N_G(\langle (1\ 2\ 3) \rangle). Lagrange’s Theorem applies. S_3
    S_3 Lemma 2 S_3
  2. G = Q_8
    H Reasoning N_G(H)
    1 Lemma 1 Q_8
    \langle -1 \rangle Lemma 3 Q_8
    \langle i \rangle We have \langle i \rangle \leq N_G(\langle i \rangle), so N_G(\langle i \rangle) is either \langle i \rangle or Q_8. Note that jij^{-1} = i^{-1}, so Lemma 6 applies, as does Lagrange’s Theorem. Q_8
    \langle j \rangle We have \langle j \rangle \leq N_G(\langle j \rangle), so N_G(\langle j \rangle) is either \langle j \rangle or Q_8. Note that kjk^{-1} = j^{-1}, so Lemma 6 applies, as does Lagrange’s Theorem. Q_8
    \langle k \rangle We have \langle k \rangle \leq N_G(\langle k \rangle), so N_G(\langle k \rangle) is either \langle k \rangle or Q_8. Note that iki^{-1} = k^{-1}, so Lemma 6 applies, as does Lagrange’s Theorem. Q_8
    Q_8 Lemma 2 Q_8
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Comments

  • Gobi Ree  On November 17, 2011 at 1:56 am

    Typo: the normalizer of \langle (1\ 2\ 3) \rangle will be S_3.

    • nbloomf  On November 17, 2011 at 10:49 am

      Thanks!

  • Gobi Ree  On November 17, 2011 at 2:03 am

    WordPress deleted my symbols. I meant that $latex $.

  • Gobi Ree  On November 17, 2011 at 2:04 am

    Omg, I don’t know how to write the subgroup generated by (123).

    • nbloomf  On November 17, 2011 at 10:51 am

      WordPress automagically interprets less-than and greater-than symbols as opening and closing an HTML tag; to subvert this, type < and >.

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