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 Post subject: GroupsPosted: Sat Aug 27, 2016 4:47 pm

Joined: Mon May 30, 2016 9:19 pm
Posts: 10
In $\mathbb{Z}$ we define $a*b=a$ . Prove that $(\mathbb{Z},*)$ isn't a group using the definition.

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 Post subject: Re: GroupsPosted: Sat Aug 27, 2016 9:26 pm
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Joined: Mon Nov 09, 2015 1:36 am
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Location: Ioannina, Greece
Assume that exists a zero element, say $e$. Then for every $a\in\mathbb{Z}$ must hold $a\ast e=a=e\ast a$. So, if we choose an element $b\neq e$ the same must hold, i.e. $b\ast e=b=e\ast b$. But then, by definition, must hold $b=b=e$. Contradiction. So, the operation $\ast$ does not define a group's structure on $\mathbb{Z}$.

P.S. By this specific operation, in any set with more than one elements we can not have a group's structure. It can be proved using the same argument.

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Grigorios Kostakos

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