# Prove that if $(a_1, a_2)$ and $(t_1, t_2)$ both belong to $R$, then $(a_1*t_1,a_2*t_2)\in R$, and $(a_1^{-1},a_2^{-1})\in R$.

Note that it is assumed that $N$ is a normal subgroup of $L$ and $R$ is the equivalence relation on $L$. That is the equivalence class $[l] = l*N:=\{l*n:n \in N\}$ of $l$ in $L$. Where the set $l*N$ is called the left $N$-coset of $l$.

Definition of $R$: $$R:= \{(l,n)\in L\times L \mid l^{-1}*n\in N\}$$ Additional definition:

$L/N := \{[l]:l\in L\}$, this is the set of equivalence classes for $R$.

The above is assumed because I have already proven it. Need help with the stated question in the title.

• I've fixed your post to make it look a little bit better. Just so you know, you can get the $\in$ symbol by typing '\in' in your formula. You can also get some MathJax tips here : math.meta.stackexchange.com/questions/5020/… Commented Sep 6, 2017 at 14:26
• Will keep that in mind thanks. Commented Sep 6, 2017 at 14:28

You want to show that $(x,y)=(a_1t_1,a_2t_2)\in R$ (I omit the $*$ symbol everywhere). So take $x^{-1}y$:

$$x^{-1}y=(a_1t_1)^{-1}a_2t_2=t_1^{-1}a_1^{-1}a_2t_2$$

Now $a_1^{-1}a_2\in N$ since $(a_1,a_2)\in R$ thus

$$t_1^{-1}a_1^{-1}a_2t_2\in t_1^{-1}Nt_2$$

This set can be further rewritten:

$$t_1^{-1}Nt_2 =t_1^{-1}N t_{1} t_{1}^{-1}t_2=Nt_1^{-1}t_2$$

The last equality follows since $N$ is normal. And since $t_1^{-1}t_2\in N$ (because $(t_1,t_2)\in R$) then $Nt_1^{-1}t_2=N$.

Therefore $x^{-1}y\in N$ which completes the proof.

For the second part consider $(x,y)=(a_1^{-1}, a_2^{-1})$ and take $x^{-1}y$, i.e.

$$(a_1^{-1})^{-1}a_2^{-1}=a_1a_2^{-1}=(a_2^{-1}a_1)^{-1}$$

Now note that since $R$ is an equivalence relation then $a_2^{-1}a_1\in N$ and thus $(a_2^{-1}a_1)^{-1}\in N$ since $N$ is a subgroup.

Therefore $x^{-1}y\in N$ which completes the proof.

• Your first part makes complete sense, but one tiny silly question. If something is an element of N in this question then it is automatically an element of R too due to R's definition correct? Commented Sep 6, 2017 at 14:27
• @T.Liment $R$ is a relation, i.e. $R\subseteq L\times L$ and $N\subseteq L$ so as you can see nothing can be in $R$ and $N$ at the same time. These are related by "$(x,y)\in R$ (or $x R y$) if and only if $x^{-1}y\in N$" (as you defined yourself). Commented Sep 6, 2017 at 14:28
• Oh okay yeah that makes sense. Cheers for the quick reply. In hindsight that was actually not that difficult. Commented Sep 6, 2017 at 14:30