Is a subgroup of a topological group a topological group? I'm trying to solve the problem from Munkres: Let $H$ be a subspace of $G$. (Where $G$ is a topological group). Show that if $H$ is a subgroup of $G$, then both H and H closure are topological groups. 
Is it enough to say that the maps $H\times H \rightarrow H$ sending $x \times y$ to $xy$ and $x \rightarrow x^{-1}$ are continuous as restrictions of the corresponding functions on $G$ as a topological group? And similarly for $H$ closure?
 A: If $H\le G$ ($H$ is a subgroup of $G$), then the restriction of $*:G\times G\to G$ to $H×H$ is continuous, being the restriction of a continuous map to a subspace (A subtlety which should be appreciated here is  that $H×H$ with the product topology is the same as $H×H⊆G×G$ with the subspace topology. Edit: I just realized you only need continuity of $H \times H \to G \times G$, the product of the inclusion $H \to G$ with itself, in order to deduce that $H \times H \to H$ is continuous). The same holds for $()^{-1}:H\to H$.
For the $\overline H$ you need to show that this closure is still a subgroup, i.e. that multiplication of two elements in $\overline H$ does not lead outside of $\overline H$. In other words, show that $*\left(\overline H×\overline H\right)\subseteq\overline H$. The equality $\overline H×\overline H=\overline{H×H}$ may be useful here.
A: $F \subseteq G$ is a subgroup of G if and only if $a,b \in F$ $\implies$ $ab^{-1}$ is an element of F. 
Another way to see this is to prove that the function $f(x,y) = x*y^{-1}$ applied to F is a subset of F. 
So we want to show that $f(\overline H×\overline H)$ $\subseteq\overline H$. 
We then simply observe f$(\overline H×\overline H)$ $=$ $f(\overline{H \times H})$ $\subseteq \overline {f(H \times H)} \subseteq \overline H$.
Thus $\overline H$ is a subgroup of $G$ and its corresponding multiplication and inversion maps are continuous as restrictions of the corresponding maps of $G$. So $\overline H$ is a topological group.   
A: In general, closed subgroups of a topological group are topological group*. Although  the maps $p:H\times H\longrightarrow G$  and $i:H\longrightarrow G$ are continuos, this not implies in general that $p:H\times H\longrightarrow H$ and $i:H\longrightarrow H$ are  continuos. 
In order to proof the clain *, we will use the following characterization of continuos function: $f:X\longrightarrow Y $ is continuous if, and only if, $f(\overline{A})\subset \overline{f(A)}$ for all subset $A$ of $X$ (Munkres, page 104). Let $A\subset H\times H$. Then, as $H\times H$ is closed in $G\times G$, the clousure of $A$ in $H\times H$ is the same as the clousure of  $A$ in $G\times G$. As $p:G\times G\longrightarrow G$ is continuos, we have that $p(\overline{A}^{H\times H})=p(\overline{A}^{G\times G})\subset\overline{p(A)}^G=\overline{p(A)}^H$, concluiding that the product operation of $H$ is continuos. As the inversion $i:G\longrightarrow G$ is continuos, we have $i(\overline A^H)=i(\overline{A}^G)\subset \overline{i(A)}^G=\overline{i(A)}^H$, concluiding that the inversion operation of $H$ is continuos. Therefore, $H$ is a topological group (here, $\overline{A}^X$ denote the closure of $A$ in $X$. Note that if $A$ and $B$ are two closed subsets of $X$ and $A\subset B$, then $A$ is closed in $B$, since that $B-A=(X-A)\cap B$). 
