$H \cong K \times H$ implies $K$ is trivial?

Question

Let $H$, $K$ be groups, and suppose that $H \cong K \times H$. Does it necessarily follow that $K$ is trivial?

After answering I noticed that this duplicates at least two earlier questions; the other one that I found is here. — Brian M. Scott, Sep 08 '15 at 05:22

score 1 · Answer 1 · answered Sep 08 '15 at 05:18

1

No: let $K$ be any group, and let $H$ be the Cartesian product of infinitely many copies of $K$.

answered Sep 08 '15 at 05:18

Brian M. Scott

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score 1 · Answer 2 · answered Sep 08 '15 at 05:19

No, there are some pretty weird spaces out there that satisfy things like $$ H\equiv H\oplus H$$ For example the sequence space $\ell^2$ (taken as an additive group) can be written as a direct sum of itself twice. You just interleaved even and odd indexes.

$H \cong K \times H$ implies $K$ is trivial?

2 Answers2