How to show that the intersection of all neighbourhood of $0$ in a topological group is a subgroup?

Question

Let $H$ be the intersection of all neighbourhood of $0$ in a topological group $G$. How to show that $H$ is a subgroup?

I tried to use the continuity of multiplication and inverse. But not successful. Thank you very much.

Edit: it seems that we need the condition that $G$ is abelian. If $G$ is not ablian, do we have $H$ is a subgroup?

Answer 1

Let $U$ be any neighbourhood of $0$, then $\exists$ a neighbourhood $V$ of $0$ such that $VV^{-1} \subset U$ : Let $f: G\times G \to G$ be the map $(a,b) \mapsto ab^{-1}$. Then, $f^{-1}(U)$ is an open neighbourhood of $(0,0)$. Hence it contains a neighbourhood $V_1\times V_2$ of $(0,0)$. Now $V = V_1\cap V_2$ works.

Let $x,y\in H$, then for any neighbourhood $U$ containing $0$, choose $V$ as above. Then $x,y \in V$, so $xy^{-1} \in U$, and so $xy^{-1} \in H$.

Thus, $H$ is a subgroup.

Answer 2

You're quite right, you do need the condition G is Abelian for this particular claim. Why isn't obvious,but your question's been asked before on here. Here's the answer and several versions of the proof.

Intersection of all neighborhoods of zero is a subgroup

It's worth noting this turns out to be a special case of a more general result:

Theorem: If G is a topological group and $H$ is a subgroup of $G$, then the closure of $H$ in the topology $\bar {H}$ is also a subgroup in $G$.

Proof. Let $g,h \in\bar {H}$. Let U be an open neighborhood of the product $gh$. Let $\mu:G \times G \rightarrow$ G denote the multiplication map, which is continuous, so $\mu^{-1}(U)$ is open in GxG, and contains $(g,h)$. So, there are open neighborhoods $V_1$ of $g$ and $V_2$ of $h$ such that $V_1 \times V_2\in\mu^{-1}(U)$. Since $g,h\in\bar {H}$, then there are points $x\in V_1\cap H \neq \emptyset$ and $y\in V_2\cap H \neq \emptyset$. Thus, $x,y\in H$, we have $xy\in H$ and since $(x,y)\in \mu^{-1}(U)$ then $xy \in U$ Therefore $xy\in U\cap H$. Since $U$ was an arbitrary open neighborhood of gh, then we have $gh\in \bar{H}$. Now let $i:G \rightarrow G$ denote the inverse map, and let W be an open neighborhood of $h ^{1}$. Then $i^{-1}(W) = W^{-1}$ is open and contains h, so there is a point $z\in H\cap W^ {1}\neq \emptyset$. But this means $z^{-1}\in H\cap W\neq\emptyset$ and it implies $h^{-1}\in\bar{H}$. But this means $\bar{H}$ is a topological subgroup of G and we're done!

Note this result is perfectly general and covers the earlier claim as a special case. So the Abelian requirement isn't really needed!

Answer 3

Let $a,b$ be in that intersection $H$. Then they are in every neighborhood of $e$. So you can easily check, that every open neighborhood of $a$ or $b$ is also a neighborhood of $e$ and vice versa (o/w apply map say $a^{-1}$). Similar you check that neighborhoods of $ab$ correspond to neighborhoods of $a$ (push everything back and forth by multiplication maps).