Show for each $c$, $\{x + c \Bbb Z \in \Bbb Z/c\Bbb Z\mid x^{c−1} \equiv 1 \pmod c\}$ is an abelian group under multiplication of congruence classes Show for each $c$, the set 
$$\{x + c \Bbb Z \in \Bbb Z/c\Bbb Z\mid x^{c−1} \equiv 1 \pmod c\}$$ is an abelian group 
under multiplication of congruence classes.
 A: $\newcommand{\Z}{\mathbb{Z}}$
Let's call the set you've defined $G$:
$$G = \left\{ x+ c\Z\mid x^{c-1}\equiv 1\pmod{c}\right\}.$$
To show that $G$ is a group, we need to ensure that it is not empty, that it is closed under multiplication, and that each element has a multiplicative inverse.  That it is associative and commutative follows because these properties are inherited from the multiplication in the ring $\Z/c\Z$.
First, to show that $G$ is non-empty, note that $1+c\Z$ belongs to $G$ because (putting $x=1$ in the definition) we have $1^{c-1} = 1\pmod{c}$.  Furthermore, this element $1+c\Z$ serves as an identity element for the set $G$ since, for any element $x+c\Z\in G$, we have
$$(1+c\Z)(x+c\Z) = 1\cdot x + c\Z = x+c\Z,$$
and we've already noted that the multiplication is commutative, so in fact, $1+c\Z$ is a two-sided identity.
Next, choose two elements $x+c\Z$ and $y+c\Z$ in $G$.  Their product (in the ring $\Z/c\Z$) is the coset $xy+c\Z$.  Our task is to show that $xy+c\Z$ is in $G$.  Now, because $x+c\Z$ belongs to $G$, we know that $x^{c-1}\equiv 1\pmod{c}$.  Likewise, $y^{c-1}\equiv 1\pmod{c}$.  But
$$(xy)^{c-1} = x^{c-1}y^{c-1}\equiv 1\cdot 1 = 1\pmod{c},$$
using basic arithmetic of congruences.  Therefore, the product $(x+c\Z)(y+c\Z) = xy+c\Z$ is in $G$, as required.
Finally, we need to demonstrate that every member of $G$ has a multiplicative inverse in $G$.  To this end, let $x+c\Z\in G$ be arbitrary; we seek an element $y+c\Z\in G$ such that the product of $x+c\Z$ and $y+c\Z$ is the identity element of $G$, which we established earlier is the coset $1+c\Z$.  This is a bit like saying that $xy = 1$ up to some squishiness and, for this, we can take note of the fact that, since $x+c\Z$ is in $G$, we must have $x^{c-1}\equiv 1\pmod{c}$.  That's a bit like saying that $x^{c-1} = 1$, up to some squishiness, so why not guess that $y = x^{c-2}$ might work?  For this choice of $y$ we at least have that
$$(x+c\Z)(y+c\Z) = (x+c\Z)(x^{c-2}+c\Z) = x\cdot x^{c-2} +c\Z= x^{c-1}+c\Z = 1+c\Z .$$
[Why?  Well, remember that $x^{c-1}\equiv 1\pmod{c}$ means that $x^{c-1}-1\in c\Z$, by definition.  (Something lives inside $c\Z$ precisely when it is divisible by $c$.) But two cosets, say, $u+c\Z$ and $v+c\Z$, are equal precisely when $u-v\in cZ$ or, in other words, $u\equiv v\pmod{c}$. Applying this (say, with $u=x^{c-1}$ and $v=1$) we see that $x^{c-1}+c\Z = 1+c\Z$.] Now, the last little detail is that we need to ensure that this $y+c\Z = x^{c-1}+c\Z$ actually lives in $G$!  But this just means checking that $y^{c-1}\equiv 1\pmod{c}$, and for this we just compute, thus:
$$y^{c-1} = (x^{c-2})^{c-1} = (x^{c-1})^{c-2}\equiv 1^{c-1}=1\pmod{c}.$$
In summary: We have a non-empty set with a multiplicative identity element ($1+c\Z$); the product of any two elements belongs again to this set; and, each element of the set has a multiplicative inverse that belongs to the set.  Commutatively and associativity of multiplication are inherited from the containing ring $\Z/c\Z$.  Therefore, this set is an Abelian group.
