Proving that the identity $e$ of $G$ is in the subgroup $H_a$ [duplicate]

I have the given group, $H_a=\{x\in G \mid xa=ax \}$, and I want to prove it is a subgroup of $G$.

The associative law of $ax=xa$ proves that the binary operation of $G$ is closed in $H_a$. But when proving that the identity $e$ of $G$ is in $H_a$, I do $a\cdot0=0$. But how do I know that $0\in G$?

Thanks

Solution 1:

Fix $a\in G$.

Use the one-step subgroup test.

We have $ea=a=ae$ by definition of $e$. Thus $e\in H_a$, so $H_a\neq\varnothing$.

By definition, $H_a\subseteq G$.

Let $x,y\in H_a$. Then we have $xa=ax$ and $ya=ay$; in particular, we have $ay^{-1}=y^{-1}a$ by multiplying on both sides of $ya=ay$ by $y^{-1}$. Now

$$\begin{align} (xy^{-1})a&=x(y^{-1}a)\\ &=x(ay^{-1})\\ &=(xa)y^{-1}\\ &=(ax)y^{-1}\\ &=a(xy^{-1}), \end{align}$$

so $xy^{-1}\in H_a$.

Hence $H_a\le G$.

Solution 2:

The identity element satisfies $ae=ea$ and so belongs to $H_a$.

There is no absorbing element in a group such as $0$ with $a0=0$.