# 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?

• That may be fine for $H$, but the "problem" with $\overline H$ is rather to show that it is a group Commented Feb 23, 2014 at 22:30
• Okay, but for the continuous functions part my argument is valid? Commented Feb 23, 2014 at 22:35

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.

• Thanks very much! I see it now. Commented Feb 23, 2014 at 23:03
• How do we show that this is a topology generated by subspace topology? Do we need to consider $U$ a open subset of $G$ and consider $U \cap H$ and show finite intersection and arbitrary union properties?
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Commented Mar 21, 2017 at 16:58
• Are you asking how we show that the product topology on $H\times H$ is the same as the subspace topology from $G\times G$? @Sayantani Commented Mar 21, 2017 at 17:48
• @Sayantani: You need to be more specific about which step of my proof you have problems with. Is it about why $*: H \times H \to H$ is continuous? This map is continuous because the composition $H \times H \stackrel*\to H \to G$ is continuous, the latter map being the inclusion of $H$ into $G$. The reason for this is that we can write the composition as $H \times H \to G \times G \stackrel*\to G$, where the first map is the square of the inclusion of $H$ into $G$. Commented Mar 21, 2017 at 18:32
• How to show when $H$ becomes a topo space in itself, it gets the subspace topology generated on $H$ as a subspace of $G$ ? Do we need to find the open sets of the subspace topology as $U \cap H$ where $U$ is open in $G$ and show that those are included in topology generated on $H$, and vice versa? I hope I could make it clear.
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Commented Mar 22, 2017 at 1:09

$F \subseteq G$ 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.

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 $$B-A=(X-A)\cap B$$).