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Let $G$ be a set with an operation $\ast:G\times G \rightarrow G$ such that:

(1) For all $a,b,c \in G$ we have $(a\ast b)\ast c=a\ast (b\ast c)$ (associativity),

(2) There is $e \in G$ such that $e\ast a=a$ for all $a\in G$ (left identity),

(3) For all $a \in G$ there is $a^{\prime}\in G$ such that $a\ast a^{\prime}=e$ (right inverse).

Show that $(G,\ast )$ need not be a group.

I can't think of a counter example that satisfies this not being a group.

user1729
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Sam Houston
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  • I think that the following is a really good start: http://math.stackexchange.com/questions/507279/example-of-left-and-right-inverse-functions – Squirtle Oct 13 '14 at 19:17

1 Answers1

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Take e.g. $G = \{0, 1\}$ with $a, b \in G: a*b = b$. Then:
- $*$ is associative
- $b \in G: 1*b = b$, thus $1$ is a left identity
- $b \in G: b * 1 = 1$, $1$ is always the right inverse

But this is obviously not a group

MJD
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agb
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