Let $H$ be a subgroup of $G$, and $x \in N_G(H)$. Prove that $$P = \{hx^i | h\in H, i\in\Bbb Z\}$$ is a subgroup of $G$.

I proved that $P$ is not empty because $H$ is a subgroup of $G$ so it's not empty and $N_G(H)$ is also not empty because $1$ commutes with every element in $H$.

I don't know how to show that $P$ is closed under $G$ operation and is closed under inverses.

To prove the former, I tried: Let $s$ and $t \in P$ so $s = h_1(x_1)^i$ and $t = h_2(x_2)^j$, so $st = h_1(x_1)^ih_2(x_2)^j$ but don't know how to show that belongs to $P$ also...

I would really appreciate any help.


2 Answers 2


We have that every power $x^n$ normalizes $H$. Thus $Hx^n=x^nH$. Hence $(h_1x^i)(h_2x^j)=h_1h_3x^{i+j}=hx^{i+j}$, $h\in H$, so $P$ is closed under multiplication. Then $(hx^i)^{-1}=x^{-i}h^{-1}=h_1x^{-i}$. So $P$ is closed under taking inverses.


There is a somewhat more general result of group theory at play here in this context and I would like to mention it explicitly. Recall that given any magma $(A, \cdot)$ -- i.e. a pair consisting of set $A$ equipped with an arbitrary binary operation $\cdot \colon A \times A \to A$ -- for any subsets $X, Y \subseteq A$ we define their subset product as $XY\colon=\{xy\}_{\substack{x \in X \\ y \in Y}}$. This allows us to consider the extension of the original operation "$\cdot$" on $A$ to the powerset $\mathscr{P}(A)$, as follows: $$\begin{align} \cdot \colon \mathscr{P}(A) \times \mathscr{P}(A) &\to \mathscr{P}(A)\\ \cdot(X, Y)\colon&=XY \end{align}$$ If we equip the powerset with the standard order structure given by inclusion, the magma structure introduced above will be compatible with this order structure (i.e. will be an ordered magma) in the precise sense that for any subsets $X, Y, Z \subseteq A$ such that $X \subseteq Y$ the relations $XZ \subseteq YZ$ respectively $ZX \subseteq ZY$ hold.

If $(S, \cdot)$ is a semigroup -- i.e. the operation "$\cdot$" on $S$ is associative -- the extension to $\mathscr{P}(S)$ remains associative, in other words the induced structure $(\mathscr{P}(S), \cdot)$ will also be a semigroup.

Proposition. Let $G$ be an arbitrary group and $K, H \leqslant G$ be two subgroups. Then the subset product $KH \leqslant G$ is a subgroup if and only if the commutation relation $KH=HK$ holds.

Proof. In groups it is in general valid that $(XY)^{-1}=Y^{-1}X^{-1}$ for any subsets $X, Y \subseteq G$. Under the hypothesis that $KH \leqslant G$, since subgroups are equal to their own symmetrics (the symmetric of subset $X \subseteq G$ is $X^{-1}\colon=\left\{x^{-1}\right\}_{x \in X}$), we infer that $KH=\left(KH\right)^{-1}=H^{-1}K^{-1}=HK$.

Conversely, assuming the commutation relation $F\colon=KH=HK$ we have the relation $FF^{-1}=(KH)(KH)^{-1}=(KH)\left(H^{-1}K^{-1}\right)=K\left(HH^{-1}\right)K^{-1} \subseteq KHK^{-1}=HKK^{-1} \subseteq HK=F$, where we have appealed to the criterion according to which $E \subseteq G$ is a subgroup if and only if $E \neq \varnothing$ and $EE^{-1} \subseteq E$. Since $K, H \neq \varnothing$ we clearly have $F \neq \varnothing$ and thus the mentioned criterion applies to $F$ as well, to the effect that $F \leqslant G$.

The proposition applies immediately to your problem: given $H \leqslant G$ and $t \in \mathrm{N}_G(H)$ we have by definition that $L\colon=\{xt^n\}_{\substack{x \in H\\n \in \mathbb{Z}}}=H\langle t \rangle$, where $\langle X \rangle$ denotes the subgroup generated by subset $X \subseteq G$ (with the abbreviation $\langle \{y\} \rangle=\langle y \rangle$ for any $y \in G$).

As $t \in \mathrm{N}_G(H)$ it follows that $\langle t \rangle=\{t^n\}_{n \in \mathbb{Z}} \leqslant \mathrm{N}_G(H)$ and one can easily prove that more generally for any subgroup $K \leqslant \mathrm{N}_G(H)$ the relation $KH=HK$ is valid. Indeed, for any $y \in K$ we have by definition of the normaliser that $yH=Hy$ and thus $KH=\displaystyle\bigcup_{y \in K}yH=\displaystyle\bigcup_{y \in K}Hy=HK$.


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