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This is possible with abelian groups for any $n$; see this answer to a very similar question.

If you require the group structure to be continuous, this is impossible. Indeed, in that case, the image $[-1,1]$ would have to be a group as well. But $[-1,1]$ is not homogeneous, so it cannot be a …

This follows fairly straightforwardly from the fact that every map from $A=\mathbb{Z}^\mathbb{N}$ to $\mathbb{Z}$ factors through a finite subproduct (let me call this "Specker's theorem"). Suppose $ …

Since $\mathbb{C}^\times \cong \mathbb{R}\_{+}\times S^1$ by polar coordinates, it suffices to show that $\text{Hom}({\mathbb R}\_+,{\mathbb R}\_+)$ is uncountable. But for any real number $a$, $x\ma …

In the abelian case, elements of infinite order have to exist, for any compact group. Let G(n) be the (closed) subgroup of n-torsion elements. If G were the union of the G(n), then by the Baire cate …

Here's a counterexample: on $\mathbb{Z}^2$, $f(x,y)=(x,y+x)$. More generally, the order-preserving automorphisms of $\mathbb{Z}^n$ are exactly the upper triangular matrices with 1s on the diagonal ( …

Spheres are (homotopy) cogroups for the same reason that homotopy groups are groups. The comultiplication $S^n \to S^n \vee S^n$ is the map that collapses the equator, the same map that is used to de …

All cohomology in this answer will have $\mathbb{Z}/2$ coefficients, and $K_n$ will denote $B^n(\mathbb{Z}/2)$. By the Yoneda lemma, the cohomology $H^m(K_n)$ can also be thought of as the natural op …

No, this isn't even true if $G$ is $S_n$ itself: the symmetric group $S_6$ has an outer automorphism.

I don't have anything to say about specific examples, but here are some general remarks. A way to construct the abelianization of any compact group is to consider its image under the product of all i …