Given groups $G_1, G_2$, if $K$ is a subgroup of $G_1 \times G_2$, are there $H_1\leq G_1$ and $H_2\leq G_2$ such that $K=H_1\times H_2$?

Asked Apr 18 '22 at 21:49

Active Apr 18 '22 at 22:27

Viewed 37 times

Duplicate Question Isn't a duplicate, I didn't mention they are finite.

Question:

Given groups $G_1, G_2$, if $K$ is a subgroup of $G_1 \times G_2$, are there $H_1\leq G_1$ and $H_2\leq G_2$ such that $K=H_1\times H_2$?

I have a strong belief that the upper claim is wrong, but as much as I try I couldn't find a valid counter example.

I have also tested the very basic case with $K=\{(e_1,e_2)\}$

edited Apr 18 '22 at 22:27

asked Apr 18 '22 at 21:49

Algo

See: https://math.stackexchange.com/questions/4212538/when-must-the-subgroup-of-a-product-be-the-product-of-subgroups – wormram Apr 18 '22 at 21:56
@MichaelMorrow I don't understand why is that a counterexample, how to prove there is so Hi so that K=H1 x H2? – Algo Apr 18 '22 at 22:26
you can't, because it's not true – wormram Apr 18 '22 at 23:30
For the case of $G_1 = G_2$, take $K$ to be the diagonal consisting of all $(g,g)$ for $g$ in $G$. – Ned Apr 18 '22 at 23:31

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