Modulo groups and non-prime numbers Let $M$ be a non prime number and $G$ be the set of non-zero integers modulo $M$, under multiplication modulo $M$.
Show this is not a group.
My attempt:
Since $M$ is non prime so there exists integers $r,s>1$ such that $rs=M$,
that is $rs=0\mod M$, so there exists integers in $G$ where closure doesn't hold. Is this correct?
Also can a left coset equal right coset if group is not abelian?
Thanks in advance.
 A: There is more than one definition of a group, which I guess is where the confusion in the comments lie. (A similar confusion came up here before, in the comments to this question.) The first definition I learned as an undergrad, which I guess is the one the OP is working with, is roughly as follows (this is an amended version of the Wikipedia definition):
A group is a set, $G$, together with an operation $\circ$ which may be applied to pairs of elements in $G$. To qualify as a group, the set and operation, $(G, \circ)$, must satisfy four requirements known as the group axioms:


*

*Closure. For all $a, b\in G$ their product $a \circ b$ is an element of $G$.

*Associativity. For all $a, b, c \in G$ we have $(a \circ b) \circ c = a \circ (b \circ c)$.

*Identity element. There exists an element $e \in G$ such that, for every element $a \in G$, the equation $e \circ a = a \circ e = a$ holds.

*Inverse element. For each $a \in G$, there exists an element $b \in G$ such that $a \circ b = b \circ a = e$, where $e$ is an (the) identity element.


As you point out, $G$ contains two (equivalence classes) of integers such that their product is not in $G$. For example, if $m=6$ then $G=\{1,2,3,4,5\}$ but $2\times3=6\not\in G$. Hence, $G$ is not a group as it fails the "closure" axiom.
(In conclusion: your proof is fine.)

You also ask:

Can a left coset equal right coset if group is not abelian?

Yes. A subgroup $H$ is normal precisely when its left and right cosets are equal. Non-abelian groups can have normal subgroups. For example, $S_3$ is non-abelian, and if $H=\{1, (123)(132)\}$ then the left cosets are $H$ and $(12)H$, while the right cosets are $H$ and $H(12)$.
In the above example, $H$ is abelian but $G=S_3$ is not. There are other examples where both $H$ and $G$ are non-abelian. For example, if you have come across the alternating group $A_n$, then $A_n$ is a normal subgroup of $S_n$ (so left cosets=right cosets), and $A_n$ is non-abelian when $n>3$. (In the above example we actually have $H=A_3$, and this is cyclic so abelian.)
