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 $G/H \cong Z_2 \times Z_4$.