• #todo/school : Section 2.2: 2, 3, 6, 11 📅 2025-01-29 ✅ 2025-01-29

2

Proof

$(\subseteq )$: Let $g\in C_G(Z(G))$. Then since $C_G(A)\subseteq A$ for any set $A$, then $g\in G$.

$(\supseteq )$: Let $g\in G$. We want to show that $\mathrm{\forall }a\in Z(G)$ that $ag=ga$, so then that $\mathrm{\forall }b\in G$ then $ab=ba$.

Now, let $a\in Z(G)$. Then $\mathrm{\forall }b\in G$ then $ab=ba$. Now since $g\in G$ then $ag=ga$ as desired. Thus then $g\in C_G(Z(G))$.

Now to deduce $N_G(Z(G))=G$, notice that for any set $A$ that:

so using our first part automatically gives us $N_G(Z(G))=G$.

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3

Proof

Let's just assume that $C_G(A)$ is a group as it's trivial to show. It is good to note that $C_G(B)\subseteq C_G(A)$ as if we let $g\in C_G(B)$ then $\mathrm{\forall }b\in B$ then $gb=ba$. Let $a\in A$ (and thus $a\in B$) so then $ga=ag$ and thus $g\in C_G(A)$.

(Closure under binary operation) Let $g_1,g_2\in C_G(B)$. Then $\mathrm{\forall }b\in B$ then $g_{1}b=b{g}_1$ and the same for $g_2$. Now let $a\in A$ and consider $g_1{g}_2$. We want to show:

To do this, notice that $a\in B$ as $A\subseteq B$, so then $g_{1}a=a{g}_1$ and the same for $g_2$. Thus:

Thus $g_1{g}_2\in C_G(A)$.

(Inverses are in the set). Let $g\in C_G(B)$. Again $\mathrm{\forall }b\in B$ then $gb=bg$. We want to show that $g^{-1}\in C_G(B)$. Now to do this let $b\in B$. Then $gb=bg$ as described before. We really want to show that $g^{-1}b=b{g}^{-1}$ but that's easy since:

as desired.

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6

Proof

(a): Let's assume that $N_G(H)$ be a group. Let's show that $H\subseteq N_G(H)$. Let $h\in H$. We want to show that $hH{h}^{-1}=H$ (because then by definition we have $h\in N_G(H)$). To do this consider the following subset arguments:

• $(\subseteq )$: Let $h{h}^{\prime }{h}^{-1}\in hH{h}^{-1}$. Then because $H$ is closed under binary operations then $h{h}^{\prime }{h}^{-1}\in H$ as desired.
• $(\supseteq )$: Let $h^{\prime }\in H$. Construct $h_0=h^{-1}{h}^{\prime }h\in H$ (by closure of $H$). Notice that:

Thus $h^{\prime }\in hHh$ as desired.

To show that this is not necessarily true if $H$ is not a subgroup, it begs to find a group that isn't a subset, as then we don't get the subset property above (and thus it's not a valid statement). Let $G=(\frac{\mathbb{Z}}{3\mathbb{Z}},+)$ and $H=(\{\bar{1},\bar{2}\},+)$. Clearly $H$ is not a valid subgroup, and $N_G(H)=G$, but $H$ still doesn't have the identity element and thus is not a subgroup here.

(b): For the forward direction, suppose $H\le C_G(H)$. Let $h_1,h_2\in H$. Then $h_1,h_2\in C_G(H)$ by $H\subseteq C_G(H)$. Then $\mathrm{\forall }{h}_0\in H(h_1{h}_0=h_0{h}_1)$. Use $h_0=h_2\in H$ to get that $h_1{h}_2=h_2{h}_1$, showing abelianness of $H$.

For the reverse direction, suppose $H$ is abelian. Then to show $H\le C_G(H)$ (similar to (a)) we only need to show $H\subseteq C_G(H)$. Let $h\in H$. To show $h\in C_G(H)$ let $h_0\in H$. Notice that by $H$ being abelian, then $h{h}_0=h_{0}h$. Thus $h\in C_G(H)$.

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11

Proof

Clearly $\star {|}_{Z(G)}$ is closed (with its inverses) so we'll only show $Z(G)\subseteq N_G(A)$ for any $A\subseteq G$.

Let $A\subseteq G$, and let $z\in Z(G)$. Then $\mathrm{\forall }g\in G(zg=gz)$ by the definition for $Z(G)$.

To show $z\in N_G(A)$ requires showing $zA{z}^{-1}=A$:

• $(\subseteq )$: Let $a^{\prime }\in zA{z}^{-1}$, so $\mathrm{\exists }a\in A$ where $a^{\prime }=za{z}^{-1}$. Now $A\subseteq G$ and $a\in A$ so then $a\in G$. Thus using $z\in Z(G)$ then $za=az$. Thus:

so then $a^{\prime }\in A$.

• $(\supseteq )$: Let $a^{\prime }\in A$. Notice then $a^{\prime }\in G$ so $z{a}^{\prime }=a^{\prime }z$. Thus $a^{\prime }=a^{\prime }z{z}^{-1}=z{a}^{\prime }{z}^{-1}\in zA{z}^{-1}$ as desired.
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