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If I have a set S under some operation, and the composite element (ab) is in S, then is it true that a,b must both be in S as well?

Could we have a situation where just a is in S but not b, or even neither being in S?

Thanks for any help. I need this question for group theory. I'm really trying to prove that if the union of two subgroups is a group then it must be true that one of them must be a subset of the other.

I was trying to prove it by contradiction, but I'm not sure if I can assume the statement above.

Andrés E. Caicedo
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Edwin Lin
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  • Say $H,K$ are two subgroups of a group $G$ such that their union $H\cup K$ is also a subgroup. If neither is a subset of the other then choose $h,k$ such that $h\in H,h\not \in K$, $k\in K, k\not \in H$. $hk $ is in the union of course hence it is in either $H$ or $K$ (or both). Can you finish from there? – lulu Sep 10 '18 at 23:09

2 Answers2

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No: take the monoid $\mathbb{N}$ under multiplication. $-2$ and $-3$ are not in $\mathbb{N}$, but their product is.

For the second situation, take the same monoid under addition and pick two integers such that only one is negative but their sum is positive.

bsbb4
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Since you need the answer for your group theory class I can give an answer about groups

Generaly: Given any group $G$ and any proper subgroup $H$ you can take some $g\in G\backslash H$ and let $a=g,b=g^{-1}$ then neither $a,b\in G$ and $ab=1\in H$

For example: If you take $G = \mathbb{Z}$, $H=2\mathbb{Z}$ and $a=1,b=-1$ then $a,b\not\in H$ but $a+b=0\in H$.

Edit: @Daniel Schepler's comment is worth reading.

also you may want to look at this question If a group is the union of two subgroups, is one subgroup the group itself?

Yanko
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    However, if $H$ is a subgroup with $ab \in H$, then $a \in H \implies b \in H$ and vice versa. So in this case, you can't have "just $a$ is in $H$ but not $b$". – Daniel Schepler Sep 11 '18 at 00:07