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Prove that any non empty set G with a binary operation $\bullet$ satisfying the following requirements is a group:

  1. If $x,y \in G$, then $x \bullet y \in G$.
  2. $\forall x,y,z \in G, (x \bullet y) \bullet z=x \bullet (y \bullet z)$
  3. $\forall a,b \in G$, the equations $x \bullet a=b$ and $a \bullet y=b$ both have solutions

Hint:

  1. Show that if $a \in G$, then a has a right and left identity and be sure to show they are the same.
  2. Show that the identity in step 1 works for any $b \in G$

$\textbf{Note:}$ Steps 1 and 2 can be reversed

  1. Show that for any $a \in G$ a has a left and right inverse and they are the same.

$\textbf{Step 1:}$ Show that if $a \in G$, then a has a right and left identity and be sure to show they are the same.

$\textbf{Definitions:}$

  • An element $e \in G$ is called a $\textbf{left identity}$ element if $e \bullet x=x$ for all $x \in G$.
  • An element $e \in G$ is called a $\textbf{right identity}$ element if $x \bullet e=x$ for all $x \in G$.

$\textbf{Proof:}$ Assume that $a \in G$. Then we need to show that (1) $e \bullet a=a$ (left identity), (2) $a \bullet e=a$ (right idenity), and (2) $e \bullet a=a=a \bullet e$

Am I reading the problem correctly? Also it seems to me odd to prove this because of the definition of identity.

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1 Answers1

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$\textbf{Definitions:}$

  • An element $e \in G$ is called a \textbf{left identity} element if $e \bullet x=x$ for all $x \in G$.
  • An element $e \in G$ is called a \textbf{right identity} element if $x \bullet e=x$ for all $x \in G$.

$\textbf{Proof:}$ We need to show that the set G with the binary operation, $\bullet$, is a group. For G to be a group with the operation $\bullet$, we must show that it has the following items: (1) Identity, (2) Inverses, (3) Associative, and (4) Closure.

From the requirements given, there is no need to prove that G is associative nor G has closure because they are given. (i.e. Requirement A implies Closure, and Requirement B implies Associativity.) So our proof will only consist of us showing that the set G with the operation $\bullet$ has (1) a unique identity element and (2) that for each element there is only one unique inverse.

$\textbf{Proof of $(G, \bullet)$ has a unique identity:}$ To prove that $(G, \bullet)$ has a unique identity we are going to use the Hint above. We don't know that e, the identity element, exists in G. By requirement 3, we know that $ax=a$ and $ya=a$ for some $x,y \in G$. So yes there is a left and right inverse, but now we need to show that x=y. Hence we must take the following cases:

  • If $ax=y$ and $1=a$, then $y=ax=1x=x$.
  • If $ax=a$ and $y=1$, then $x=1=y$.
  • If $a=a$ and $x=y$, then $x=y$.
  • If $a=y$ and $x=a$, then $x=y$.

Hence we have showed that a has an identity element. We now need to show that all elements in G will have an identity element.

Let $ b\in G$ be an arbitrary element. If we follow the proof above we will get that b will have an identity element. Since b is an arbitrary element in G, then all the elements in G have an identity element.

Note that the identity element must be unique. The proof of this claim is quite trival. Assume $e,e' \in G$ are distinct identity elements.

$$eg=ge=g \,and \,e'g=ge'=g$$

But \begin{equation*} \begin{aligned} ge=g=ge' & \iff e=e' \\ eg=g=e'g & \iff e=e' \\ \end{aligned} \end{equation*} So $e=e'$ is unique.

$\textbf{Proof that for each element in $(G, \bullet)$ there is only one unique inverse:}$ We know from the previous proof that G has an identity element, let's call e. We need to show that every element of $(G, \bullet)$ has a unique inverse. Assume $g_1,g_2 \in G$ are distinct inverse elements for an arbitrary $g \in G$.

Assume $gg_1=g_1g=e$ and $gg_2=g_2g=e$.

  • If $gg_1=e$, then $(gg_1)g_2=eg_2=g_2$.
  • If $gg_2=e$, then $g_1(gg_2)=g_1e=g_1$.

So $g_1=g_1(gg_2)=(g_1g)g_2=g_2$ is unique.

Hence $(G, \bullet)$ is a group.

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