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Problem 41 (The Identity And Inverses Are Unique)

Suppose that $(G,\cdot)$ is a group.

  1. Prove that the identity of the group is unique.
  2. Prove that if $x\in G$, then the inverse of $x$ is unique.

Solution

1. Suppose that $(G,\cdot)$ is a group. In order to prove that the identity of the group is unique it must be shown that there is in fact one and only one identity $e$ that exists in $G$.

First, by the definition of a group there is an $e\in G$ such that for all $x\in G$, $x\cdot e=e\cdot x=x$. Since $e$ exists in $G$ we must now show that there is only one identity element in $G$. Lets suppose $e_1$ and $e_2$ are both identity elements of $G$. We must show that $e_1=e_2$. First, let $x=e_1$ such that $e_1\cdot e_2=e_1$. Now let $x=e_2$ and therefore $e_1\cdot e_2=e_2$. Thus, it is shown that $e_1=e_2$.

2. Suppose that $(G,\cdot)$ is a group. In order to prove that if $x\in G$, then the inverse of $x$ is unique, we must show that there is one and only one inverse of $x$ that exists in $G$.

First, by the definition of a group for all $x\in G$ there exists a $y\in G$ such that $x\cdot y=y\cdot x=e$, where $e$ is the identity of $G$ (this is the definition of the inverse of x). Since we now know that the inverse of $x$ exists in $G$, we must show that there is only one inverse of $x$ in $G$. Let's suppose that $y_1$ and $y_2$ are both inverses of x. Then it follows that $x\cdot y_1=e$ and $x\cdot y_2=e$ which means $x\cdot y_1=x\cdot y_2$ since the identity of $G$ is unique.

Now, by multiplying on the left by $y_1$ we obtain $y_1\cdot x\cdot y_1=y_1\cdot x\cdot y_2$. Then, by the associative property $(y_1\cdot x)\cdot y_1=(y_1\cdot x)\cdot y_2$ we have $e\cdot y_1=e\cdot y_2$ and thus, $y_1=y_2$.

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