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Let $G$ be a topological group. Let $C$ be any connected component of $G$. It seems that $C$ may not be a subgroup of $G$ unless $C$ contains the identity $1$ of $G$. However, is it true that $C$ has a structure of a group on its own?

For example, let $A$ be any finitely generated abelian group. Consider the character group $\widehat{A}:=\mathrm{Hom}(A,\Bbb C^\times)$. As groups, $\widehat{A}$ is isomorphic to $F \times (\Bbb C^\times)^{r}$ with $F$ is a finite abelian and $r=\mathrm{rank}(A)$. Is it enough to say that $\widehat{A}$ is a $topological$ group which has a finite number of connected components? It seems that each component is isomorphic (as groups) to a torus $(\Bbb C^\times)^{r}$ . And is it true that each connected component itself is a group?

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    $\begingroup$ It's a coset of the connected component of the identity (which is a normal subgroup). $\endgroup$ – Lord Shark the Unknown Jan 18 '18 at 7:35
  • $\begingroup$ I am aware of that the connected component of the identity is a normal subgroup. How about an arbitrary connected component? $\endgroup$ – user Jan 18 '18 at 7:58
  • $\begingroup$ An non identity component is not a subgroup: it doesn't have an identity, and moreover is not even closed under multiplication. As a topological space it is homeomorphic, however to the identity component. $\endgroup$ – Ishan Levy Jan 18 '18 at 8:03
  • $\begingroup$ So in the example above, each component is actually homeomorphic (not isomorphic as groups) to the torus $(\mathbb{C}^\times)^r$? $\endgroup$ – user Jan 18 '18 at 8:14
  • $\begingroup$ Yes. To obtain a homeomorphism, you can choose an element $h$ of the non identity component, and multiplication by $h$ gives a continuous map from the identity component to that component, with inverse given by multiplying by $h^{-1}$. $\endgroup$ – Ishan Levy Jan 18 '18 at 8:52

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