Simple question concerning definition of the equivalence relation and the group operation in quotient group

Question

I have simple, yet critical questions for my understanding on the quotient group $G/H$.
Remember that $H$ is a sub group of $G$. So we say that, for any given element $x,y \in G$, $x$ and $y$ are equivalent (written $x\sim y$) iff $x*y^{-1} \in H$. And we write $G/H$ the set of all equivalent classes.

My questions are:

1. How can we know how the equivalent relation “$\sim$” is defined?
   For example, in $\mathbb{Z}/p\mathbb{Z}$ two numbers in $\mathbb{Z}$ are equivalent iff they have the same rest after the euclidian division by $p$.
   But in $\mathbb{R}/\mathbb{Z}$ two numbers $x,y$ in $\mathbb{R}$ are said to be equivalent iff $x-y=0$.

2. How can I know how or what is the operation noted “$*$” of the group $G$? I need to know it in order to know $g^{-1}$ (that must verify $g*g^{-1}=e$, $e$ the neutral element of the operation “$*$”)?

Is there some general convention that I should know?

Thank you

Answer 1

By Convention, $\mathbb{Z}$ as a group is equipped with addition. It makes sense because if it was multiplication, $0$ (which is in $\mathbb{Z}$) wouldn't have an inverse. Same goes for $\mathbb{R}$.

From this you can understand why the equivalence relation of $\mathbb{Z}/p\mathbb{Z}$ is $x - y \in p\mathbb{Z}$ (which is the same as $x$ and $y$ having the same rest after euclidian division by $p$), and why for $\mathbb{R}/\mathbb{Z}$, it is $x - y \in \mathbb{Z}$

Answer 2

To verify that $\sim$ is an equivalence relation, you must check the following:

• Reflexivity: For all $x \in G$, we have $x \sim x$
• Symmetry: For all $x, y \in G$, if $x \sim y$, then $y \sim x$
• Transitivity: For all $x, y, z \in G$, if $x \sim y$ and $y \sim z$, then $x \sim z$

For example, reflexivity is seen by noting that for all $x \in G$, we have $x * x^{-1} = e \in H$ since $H$ is a subgroup. Can you check the other two axioms similarly?

With regard to your question about the operation $*$, whenever people talk about a group $G$, the operation $*$ is inherently part of the data of $G$ (otherwise, $G$ would just be a set). I guess the extremely precise way to say this would be to say that $(G, *)$ is a group.

Answer 3

I think that this may help. Below a list of the main group with the operation that is associated to them.

• $(\mathbb{Z} ; +) , (\mathbb{R} ; +) , (\mathbb{C} ; +) , (\mathbb{Z}/n\mathbb{Z} ; +)$ with $+$ the casual addition.
• $(\mathbb{K}^*; \times)$ with $\mathbb{K}$ a mathematic field, $\mathbb{K}^*$ the set of all element of $\mathbb{K}$ that have an inverse with respect to the operation $ \times $ that is usually the comom multilpication. For exemple with $\mathbb{K}= \mathbb{C} \Rightarrow \mathbb{C}^* = $ the set of all complex number that are not zero.
• $ (GL_n(\mathbb{K}); \times )$ with $GL_n(\mathbb{K})$ the set of all $n×n$ invertible matrices over a field $\mathbb{K}$. This is a group with the operation of matrix multiplication "$ \times $" .
• $(S_n; \circ ) $ with $S_n$ the symetric set whose elements are all the possible bijections from the a finite set of size $n$ to itself and $\circ$ the composition of function.

Rem: More over generally as far as I know the "quotient group" $G/H$ (if it is a group and that must be checked), inherit of the operation of the group $G$.

Edit: A last precision in the case that we writte $G^*$ it implies that $G$ is a ring, and that $G^*$ is the set of all the elements invertible according to the segond law "$*$" of the ring $G$. More over $(G^*; \times)$ is a group of neutral element $1_G$.