Prove that L is a subset of M

Question

I've a question: How do I prove that:

$$L = \{a \in M \mid \forall b \in M, a * b = b * a\}$$ is a subgroup of $$\langle M; * ; (\cdot)^{-1}; e \rangle $$

I know that for a subgroup, following rules need to be fulfilled:

1) $a * b\in L$

2) $e\in L$

3) inverse of $a\in L$ for all $a\in L$.

But how do I start? I don't really see the values I've and how I can even start with proving the first "rule"...

I hope someone can help me!

Thanks in advance!!!

Greetings

Answer 1

The definition of $L$ already indicates that $L$ is a subset of $M$.

For proving that this subset is also a subgroup it is necessary and sufficient to show that: $$x,y\in L\implies xy^{-1}\in L$$

Also it is necessary and sufficient to prove that $L$ is closed under multiplication and the formation of inverse elements: $$x,y\in L\implies xy\in L\text{ and } x\in L\implies x^{-1}\in L$$

Let me show that indeed $x,y\in L\implies xy\in L$.

If for $x,y\in L$ then for every element $b$ we find that:$$(xy)b=x(yb)=x(by)=(xb)y=(bx)y=b(xy)$$

• first equality is based on associativity.
• second equality is based on $y\in L$.
• third equality is based on associativity.
• fourth equality is based on $x\in L$.
• fifth equality is based on associativity.

This gives you a start.

Answer 2

Here's a hint for the first item (closure): if $a, c\in L$, you need to show that $(a*c)*b = b*(a*c)$ for all $b\in M$. That will show that $a*c\in L$. Well, multiplication in $M$ is associative, and $a, c\in L$. Can you use those facts to show what you need to show?