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Let $G$ be a group. Let $1 \in G$ be the identity element of $G$. Let $S \subset G$, with $1 \notin S$. Is it possible for $S$ to be a group, with some other element playing the role of the identity in $S$?

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Well, $xy = x$ in a group implies $y = 1$. So it seems like the answer is no. – Dylan Moreland Jul 18 '12 at 10:16
Indeed, the identity is the only idempotent in a group (idempotent$\Leftrightarrow g^2=g$) but here you'd need two. – user1729 Jul 22 '12 at 15:30
up vote 3 down vote accepted

A subset $S$ of $G$ can not be a subgroup with respect to the multiplication in $G$ since the element $1$ is unique and is not in $S$. (I think Dylan explained this well in his comment).

If you are asking if $S$ can be a group with different multiplication than the multiplication in $G$ then take for example $S=\{g\}$ for some $g\in G$ and define $g\bigstar g=g$.

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