How to show that $J$ is a left ideal of $R$? Let $R$ be a ring and $x \in R$. Let, $J = \{ax|a \in R\}$
Show that $J$ is a left ideal of $R$.
To show something is a left Ideal(or a right ideal), do we need to show that it is a subgroup first?
Here is what I did: 


*

*$J \neq \emptyset$, since $0 \in J$

*Let, $a,b \in J$. Then, $ax - bx = (a-b)x.$ Thus, $(a-b) \in J$

*$b(ax) = (ba)x = (ab)x$. Thus, $(ab) \in J$
Am I on the right track? Is it same for right ideal?
 A: I think you are a little unclear about what elements of the ideal look like.  Every element of the ideal has the form $ax$ for some $a \in R$.  So for 2, you probably don't want $a, b \in J$, you probably want $ax, bx \in J$.  Also, your ring likely isn't commutative, so you can't assume that $ba = ab$.
A: (2) You can't conclude that $a-b\in J$ for literally any $a,b\in R$. That should be clearly impossible, as it would mean every element of the ring itself was an element of $J$ ($a-b$ will vary over every element of the ring as $a,b$ vary over element of the ring). Furthermore, you are not trying to conclude that $a-b\in J$ anyway. Closure under subtraction means if $c,d\in J$ then $c-d\in J$; in this case if $c,d\in J$ then $c=ax$ and $d=bx$ for some $a,b\in R$ (by definition of $J$), and then we can simply observe that $c-d=ax-bx=(a-b)x$ is a left-multiple of $x$, hence in $J$.
(3) Same problem as before. You cannot conclude that $ab\in J$ for arbitrary $a,b$, nor were you supposed to. We need $J$ to be closed under ambient left multiplication, which means if $c\in J$ and we have any $b\in R$, then $bc\in J$ is what we need to prove. If $c\in J$ then $c=ax$ for some $a\in R$ just as before, and we observe $bc=b(ax)=(ba)x$ is a left-multiple of $x$, hence is in $J$. Note that it is not possible to switch the order of $a$ and $b$ around, as there's no reason to assume they commute.
