Monthly Archives: February 2010

A criterion for normalcy of finite subgroups, on a generating set

Let G be a group, N \leq G a finite subgroup, and suppose N = \langle S \rangle. Prove that g \in N_G(N) if and only if gSg^{-1} \subseteq N.


(\Leftarrow) We have gSg^{-1} \subseteq gNg^{-1} = N.

(\Rightarrow) Suppose gSg^{-1} \subseteq N. Then gNg^{-1} = g\langle S \rangle g^{-1} = \langle gSg^{-1} \rangle \leq \langle N \rangle = N. By a previous theorem, g \in N_G(N).

A criterion for normalcy of finite subgroups

Let G be a group and N a finite subgroup of G. Show that gNg^{-1} \subseteq N if and only if gNg^{-1}. Deduce that N_G(N) = \{ g \in G \ |\ gNg^{-1} \subseteq N \}.


It suffices to show that for all g \in G, gNg^{-1} \subseteq N implies gNg^{-1} = N. Let g \in G. The mapping n \mapsto gng^{-1} is a bijection N \rightarrow gNg^{-1}, so that |gNg^{-1}| = |N|. Since N is finite, gNg^{-1} = N.

The last statement follows trivially.

The subgroup generated by all elements of some fixed order is normal

Let G be a group and let a,b \in G.

  1. Let g \in G. Note that g(ab)b^{-1} = gag^{-1} \cdot gbg^{-1}. Prove that |a| = |gag^{-1}|.
  2. Let g \in G. Prove that ga^{-1}g^{-1} = (gag^{-1})^{-1}.
  3. Let N = \langle S \rangle be a subgroup of G. Prove that if gSg^{-1} \subseteq N for all g \in G, then N is normal in G.
  4. Prove that a cyclic subgroup N = \langle x \rangle is normal in G if and only if for each g \in G, gx^{-1}g^{-1} = x^k for some k \in \mathbb{Z}.
  5. Let n be a positive integer. Prove that the subgroup N \leq G generated by all elements of order n is normal in G.

  1. We have g(ab)g^{-1} = ga1bg^{-1} = gag^{-1}gbg^{-1}. Now by a lemma to a previous exercise, (gag^{-1})^n = ga^ng^{-1} for all n \in \mathbb{Z}. Thus |gag^{-1}| \leq |a|. If |gag^{-1}| = k, then ga^kg^{-1} = 1, so that a^k = 1. Thus |a| \leq |gag^{-1}|.
  2. Note that gag^{-1}ga^{-1}g^{-1} = gaa^{-1}g^{-1} = gg^{-1} = 1. By the uniqueness of inverses, (gag^{-1})^{-1} = ga^{-1}g^{-1}.
  3. Let N = \langle S \rangle, and suppose gSg^{-1} for all g \in G. Then for all g \in G, we have gNg^{-1} = g \langle S \rangle g^{-1} = \langle gSg^{-1} \rangle \subseteq \langle N \rangle = N. Thus N is normal in G.
  4. Let N = \langle x \rangle. (\Leftarrow) If N is normal, then for all g \in G we have gxg^{-1} \in \langle x \rangle. Thus gxg^{-1} = x^k for some integer k. (\Rightarrow) If for all g \in G there exists an integer k such that gxg^{-1} = x^k \in N, by the previous point we have N normal in G.
  5. Let S \subseteq G consist of all elements of order n, and let s \in S. For all g \in G, by the first point we have |gsg^{-1}| = |s| = n \in \langle S \rangle. By the third point, \langle S \rangle is normal in G.

A criterion for subgroup normalcy

  1. Let G be a group and let N \leq G. Prove that N is normal in G if and only if gNg^{-1} \subseteq N for all g \in G.
  2. Let G = GL_2(\mathbb{Q}), let N be the subgroup of upper triangular matrices with integer entries and 1s on the diagonal, and let g be the diagonal matrix with entries 2, 1. Show that gNg^{-1} \subseteq N but that g does not normalize N.

  1. (1) Suppose first that N is normal. Then N_G(N) = G; thus, for all g \in G, we have gNg^{-1} = N. In particular, for all g \in G, gNg^{-1} \subseteq N. (2) Suppose gNg^{-1} \subseteq N for all g \in G. Then for all g \in G, we have N = gg^{-1}Ngg^{-1} \subseteq gNg^{-1}, so that gNg^{-1} = N. Hence N_G(N) = G.
  2. Let a = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \in N. Clearly then gag^{-1} = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \in N, so that gNg^{-1} \subseteq N. g does not normalize N, however, because no matrix in N whose (1,2) entry is odd is the g-conjugate of any element in N.

The intersection of a subgroup and a normal subgroup is subnormal

Let G be a group, and let N \leq G be normal. Prove that if H \leq G, then N \cap H \leq H is normal.


Let h \in H. Then using lemma 1 to this previous exercise, we have h(N \cap H)h^{-1} = hNh^{-1} \cap hHh^{-1} = N \cap H. Thus N \cap H is normal in H.

The join of an arbitrary collection of normal subgroups is normal

Let G be a group. Prove that the join of any nonempty collection \mathcal{H} = \{ H_i \}_{i \in I} of normal subgroups of G is normal.


Lemma: Let G be a group, g \in G, and \{ A_i \}_{i \in I} a nonempty collection of subgroups of G. Then g(\bigcup A_i)g^{-1} = \bigcup g A_i g^{-1}. Proof: \subseteq) Let x \in g(\bigcup A_i)g^{-1}. Then x = gyg^{-1} for some y \in \bigcup A_i. Thus there exists k \in I with y \in A_k, so that x \in g A_k g^{-1}. Hence x \in \bigcup g A_i g^{-1}. \supseteq Let x \in \bigcup g A_i g^{-1}. Then there exists k \in I with x \in g A_k g^{-1}, so that x = gyg^{-1} for some y \in A_k. But y \in \bigcup A_i, so that x \in g (\bigcup A_i) g^{-1}. \square

Recall that the join of \mathcal{H} is \langle \bigcup \mathcal{H} \rangle. Now let g \in G; we saw in a previous exercise that g \langle \bigcup \mathcal{H} \rangle g^{-1} = g \langle \bigcup H_i \rangle g^{-1} = \langle g(\bigcup H_i)g^{-1} \rangle = \langle \bigcup gH_ig^{-1} \rangle = \langle \bigcup H_i \rangle = \langle \bigcup \mathcal{H} \rangle. Hence \langle \bigcup \mathcal{H} \rangle is normal in G.

The set of normal subgroups of a fixed group is closed under arbitrary intersections

Let G be a group.

  1. Let H,K \leq G be normal subgroups. Prove that H \cap N \leq G is normal.
  2. Let \{H_i\}_{i \in I} be a family of subgroups H_i \leq G, all normal. Show that \bigcap H_i \leq G is normal.

  1. Using a lemma to a previous exercise, for all g \in G we have g(H \cap K)g^{-1} = gHg^{-1} \cap gKg^{-1} = H \cap K. Thus H \cap K is normal in G.
  2. Using a lemma to a previous exercise, for all g \in G we have g(\bigcap H_i)g^{-1} = \bigcap gH_ig^{-1} = \bigcap H_i. By Theorem 6 in the text, \bigcap H_i is normal in G.

Perform explicit computations in a quotient of a direct product of cyclic groups

Let G = Z_4 \times Z_4 be given by the presentation \langle x,y \ |\ x^4 = y^4 = 1, xy = yx \rangle. Note that every subgroup of G is normal, and let H = \langle x^2,y^2 \rangle = \{ 1, x^2,y^2 \}. Consider G/H.

  1. Show that |G/H| = 8.
  2. Exhibit each element of G/H in the form \overline{x}^a \overline{y}^b for some integers a and b.
  3. Find the order of each element in G/H.
  4. Prove that G/H \cong Z_4 \times Z_2.

  1. We can explicitly compute the elements of G/H as follows. G/H = \{ \overline{1} = \{ 1, x^2y^2 \}, \overline{x} = \{ x, x^3y^2 \}, \overline{x}^2 = \{ x^2, y^2 \}, \overline{x}^3 = \{ x^3, xy^2 \}, \overline{y} = \{ y, x^2y^3 \}, \overline{x}\,\overline{y} = \{ xy, x^3y^3 \}, \overline{x}^2\overline{y} = \{ x^2y, y^3 \}, \overline{x}^3\overline{y} = \{ x^3y, xy^3 \} \}.
  2. x Reasoning |x|
    \overline{1} 1
    \overline{x} 4
    \overline{x}^2 2
    \overline{x}^3 \overline{x}^3 = \overline{x}^{-1} 4
    \overline{y} \overline{y}^2 = \overline{x}^2, \overline{y}\,\overline{x}^2 = \overline{x}^2\overline{y}, \overline{y}\,\overline{x}^2\overline{y} = \overline{x}^2\overline{y}^2 = \overline{1} 4
    \overline{x}\,\overline{y} (\overline{x}\,\overline{y})^2 = \overline{x^2y^2} = \overline{1} 2
    \overline{x}^2\overline{y} (\overline{x}^2\overline{y})^2 = \overline{x}^2, \overline{x}^2\overline{y}\overline{x^2} = \overline{y}, \overline{x}^2\overline{y}\,\overline{y} = \overline{x^2y^2} = \overline{1}. 4
    \overline{x}^3\overline{y} (\overline{x}^3\overline{y})^2 = \overline{x^2y^2} = \overline{1} 2
  3. We let Z_4 \times Z_2 = \langle u,v \ |\ u^4 = v^2 = 1, uv = vu \rangle. Define a mapping \varphi : Z_4 \times Z_2 \rightarrow G/H by (u^a,v^b) \mapsto \overline{x}^a\overline{y}^b. Since G/H is abelian, \varphi is clearly a homomorphism. Moreover, since every element of G/H has the form \overline{x}^a\overline{y}^b for some 0 \leq a < 4 and 0 \leq b < 2, \varphi is surjective. Since |G/H| = 8, \varphi is bijective. Thus G/H \cong Z_4 \times Z_2.

Compute the elements and isomorphism type of a quotient group

Let G = \mathbb{Z}/(24) and H = \langle \overline{12} \rangle = \{ \overline{0}, \overline{12} \}.

  1. Compute the elements and orders of elements of G/H.
  2. Show that G/H \cong \mathbb{Z}/(12).

  1. We can easily see that G/H = \{ \overline{\overline{0}} = \{ \overline{0}, \overline{12} \}, \overline{\overline{1}} = \{ \overline{1}, \overline{13} \}, \overline{\overline{2}} = \{ \overline{2}, \overline{14} \}, \overline{\overline{3}} = \{ \overline{3}, \overline{15} \}, \overline{\overline{4}} = \{ \overline{4}, \overline{16} \}, \overline{\overline{5}} = \{ \overline{5}, \overline{17} \}, \overline{\overline{6}} = \{ \overline{6}, \overline{18} \}, \overline{\overline{7}} = \{ \overline{7}, \overline{19} \}, \overline{\overline{8}} = \{ \overline{8}, \overline{20} \}, \overline{\overline{9}} = \{ \overline{9}, \overline{21} \}, \overline{\overline{10}} = \{ \overline{10}, \overline{22} \}, \overline{\overline{11}} = \{ \overline{11}, \overline{23} \}, \}. Moreover, the orders of these elements are as follows.
    x Reasoning |x|
    \overline{\overline{0}} 1
    \overline{\overline{1}} 12
    \overline{\overline{2}} \overline{\overline{2}} + \overline{\overline{2}} = \overline{\overline{4}}, \overline{\overline{2}} + \overline{\overline{4}} = \overline{\overline{6}}, \overline{\overline{2}} + \overline{\overline{6}} = \overline{\overline{8}}, \overline{\overline{2}} + \overline{\overline{8}} = \overline{\overline{10}}, \overline{\overline{2}} + \overline{\overline{10}} = \overline{\overline{12}} = \overline{\overline{0}} 6
    \overline{\overline{3}} \overline{\overline{3}} + \overline{\overline{3}} = \overline{\overline{6}}, \overline{\overline{3}} + \overline{\overline{6}} = \overline{\overline{9}}, \overline{\overline{3}} + \overline{\overline{9}} = \overline{\overline{12}} = \overline{\overline{0}} 4
    \overline{\overline{4}} \overline{\overline{4}} + \overline{\overline{4}} = \overline{\overline{8}}, \overline{\overline{4}} + \overline{\overline{8}} = \overline{\overline{12}} = \overline{\overline{0}} 3
    \overline{\overline{5}} \overline{\overline{5}} + \overline{\overline{5}} = \overline{\overline{10}}, \overline{\overline{5}} + \overline{\overline{10}} = \overline{\overline{15}} = \overline{\overline{3}}, \overline{\overline{5}} + \overline{\overline{3}} = \overline{\overline{8}}, \overline{\overline{5}} + \overline{\overline{8}} = \overline{\overline{13}} = \overline{\overline{1}}, \overline{\overline{1}} + \overline{\overline{5}} = \overline{\overline{6}}, \overline{\overline{6}} + \overline{\overline{5}} = \overline{\overline{11}}, \overline{\overline{5}} + \overline{\overline{11}} = \overline{\overline{16}} = \overline{\overline{4}}, \overline{\overline{5}} + \overline{\overline{4}} = \overline{\overline{9}}, \overline{\overline{5}} + \overline{\overline{9}} = \overline{\overline{14}} = \overline{\overline{2}}, \overline{\overline{5}} + \overline{\overline{2}} = \overline{\overline{7}}, \overline{\overline{5}} + \overline{\overline{7}} = \overline{\overline{12}} = \overline{\overline{0}} 12
    \overline{\overline{6}} \overline{\overline{6}} + \overline{\overline{6}} = \overline{\overline{12}} = \overline{\overline{0}} 2
    \overline{\overline{7}} \overline{\overline{7}} = -\overline{\overline{5}} 12
    \overline{\overline{8}} \overline{\overline{8}} = -\overline{\overline{4}} 3
    \overline{\overline{9}} \overline{\overline{9}} = -\overline{\overline{3}} 4
    \overline{\overline{10}} \overline{\overline{10}} = -\overline{\overline{2}} 6
    \overline{\overline{11}} \overline{\overline{11}} = -\overline{\overline{1}} 12
  2. Since |\overline{\overline{1}}| = 12, G/H is cyclic. Thus G/H \cong \mathbb{Z}/(12).

Perform explicit computation in a quotient of the modular group of order 16

Let G = M = \langle u,v \ |\ u^2 = v^8 = 1, vu = uv^5 \rangle and let H = \langle v^4 \rangle = \{ 1, v^4 \}.

  1. Show that the order of G/H is 8.
  2. Exhibit each element of G/H in the form \overline{u}^a\overline{v}^b for some integers a and b.
  3. Find the order of each element of G/H.
  4. Write each of the following elements in the form \overline{u}^a \overline{b}: \overline{vu}, \overline{uv^{-2}u}, and \overline{u^{-1}v^{-1}uv}.
  5. Prove that G/H is isomorphic to Z_2 \times Z_4.

  1. We can find the elements of G/H explicitly: G/H = \{ \overline{1} = \{ 1, v^4 \}, \overline{v} = \{ v,v^5 \}, \overline{v}^2 = \{ v^2,v^6 \}, \overline{v}^3 = \{ v^3, v^7 \}, \overline{u} = \{ u, uv^4 \}, \overline{u}\,\overline{v} = \{ uv,uv^5 \}, \overline{u}\,\overline{v}^2 = \{ uv^2,uv^6 \}, \overline{u}\,\overline{v}^3 = \{ uv^3,uv^7 \} \}.
  2. x Reasoning |x|
    \overline{1} 1
    \overline{v} 4
    \overline{v}^2 (\overline{v}^2)^2 = \overline{v^4} = \overline{1} 2
    \overline{v}^3 |x^{-1}| = |x| 4
    \overline{u} 2
    \overline{u}\,\overline{v} (\overline{u}\,\overline{v})^2 = \overline{uvuv} = \overline{u^2 v^6} = \overline{v}^2, \overline{u}\,\overline{v}\overline{v}^2 = \overline{u}\,\overline{v}^3, and \overline{u}\,\overline{v}\,\overline{u}\,\overline{v}^3 \overline{uvuv^3} = \overline{u^2v^8} = \overline{1}. 4
    \overline{u}\,\overline{v}^2 (\overline{u}\,\overline{v}^2)^2 = \overline{uv^2uv^2} = \overline{u^2v^12} = \overline{v^4} = \overline{1} 2
    \overline{u}\,\overline{v}^3 |x| = |x^{-1}| 4
  3. We have \overline{vu} = \overline{uv^5} = \overline{u}\,\overline{v}, \overline{uv^{-2}u} = \overline{uv^6u} = \overline{u^2v^30} = \overline{v^6} = \overline{v}^2, and \overline{u^{-1}v^{-1}uv} = \overline{uv^7uv} = \overline{u^2v^36} = \overline{v^4} = \overline{1}.
  4. We let Z_2 = \langle x \rangle and Z_4 = \langle y \rangle. Define a mapping \varphi : Z_2 \times Z_4 \rightarrow G/H by (x^a,y^b) \mapsto \overline{u}^a\overline{v}^b. (Homomorphism) We have \varphi((x^a,y^b)(x^c,y^d)) = \varphi(x^{a+c},y^{b+d}) = \overline{u}^{a+c}\overline{v}^{b+d} = \overline{u}^a\,\overline{u}^c\,\overline{v}^b\,\overline{v}^d. Note that \overline{v}\,\overline{u} = \overline{vu} = \overline{uv^5} = \overline{u}\,\overline{v}, so we have = \overline{u}^a\,\overline{u}^c\,\overline{v}^b\,\overline{v}^d = \overline{u}^a\,\overline{v}^b\,\overline{u}^c\,\overline{v}^d \varphi(x^a,y^b)\varphi(x^c,y^d). Thus \varphi is a homomorphism. (Surjective) We saw above that every element of G/H is of the form \overline{u}^a\overline{v}^b for some 0 \leq a < 2 and 0 \leq b < 4; clearly, then, \varphi is surjective. Now we know that |G/H| = |Z_2 \times Z_4| < \infty, so that \varphi is bijective. Thus G/H \cong Z_2 \times Z_4.