0

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

Shaun
  • 47,747
Superunknown
  • 3,089

2 Answers2

1

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

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

Wuestenfux
  • 21,302
  • I am puzzled by the correct point of user1729 that $a\cdot0=a$ . I can't find this operation in my book, explained in detail. What is this operator $\cdot$ called? Eventually, if it is a regular dot product, why is it a and not 0? – Superunknown Dec 10 '21 at 16:38
  • 2
    @vqngs The operation $\cdot$ simply denotes the operation in the group $G$. This operation varies from group to group. (Confusingly, we do often refer to it as "multiplication" although it is not in general multiplication of numbers...) – user1729 Dec 10 '21 at 17:52
  • "There is no absorbing element in a group such as $0$ with $a0=0$" Well, unless $G={0}$ . . . – Shaun Dec 10 '21 at 18:29
  • @user1729 Thanks. Does it have a name? – Superunknown Dec 11 '21 at 09:49
  • The centralizer of a subset S of group G is defined as $C_G(S) = {g\in G\mid gs=sg\forall s\in S}$. – Wuestenfux Dec 11 '21 at 12:55
  • @vqngs Sort of. Groups are abstract, and we often call the abstract operation "multiplication". For example, we might say "when we multiply an element $g$ by the identity $e$ we get $g$ again", and we write $g\cdot e=g$. However, the operation in the group could be anything, and is not necessarily multiplication of numbers. – user1729 Dec 11 '21 at 18:33
  • @user1729 . Thanks! – Superunknown Dec 13 '21 at 09:04
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 $H_a\le G$.

Shaun
  • 47,747