I have a doubt regarding the group {4,8,12,16}under multiplication modulo 20. I faced a question show that {4,8,12,16} is a group under multiplication mod 20.It is fine.I have solved the problem also.But I am feeling something strange in it.The idenity element of multiplication modulo 20 should be 1 which is not in the group.I mean does the identity element change it I change the set I am dealing with keeping the binary operation same.Please give me a similiar example in which identity element under the same operation changes due to changing the set in a group.
 A: The following is a more advanced way of looking at this. Right now, this may not be at the right level for the OP - building the Cayley table is probably the right way to go for the OP - but it provides insight into what is actually going on.
Look at the ring ${\mathbb Z}/20$.
This is isomorphic to ${\mathbb Z}/4 \times {\mathbb Z}/5$;
the isomorphism from the latter to the former is given by $(x,y) \mapsto 5x - 4y$.
Now look at all elements of ${\mathbb Z}/4 \times {\mathbb Z}/5$ of the form $(0,x)$
with $x \in {\mathbb Z}/5^*$, i.e., at $\{0\} \times {\mathbb Z}/5^*$. Because $0^2=0$,
this forms a group under multiplication even though $0$ is not invertible in ${\mathbb Z}/4$.
The elements $(0,1),(0,2),(0,3),(0,4)$ of ${\mathbb Z}/4 \times {\mathbb Z}/5$ correspond to
$-4=16$, $-8=8$, $-12=4$, $-16=4$ of ${\mathbb Z}/20$, so that is exactly the OP's example. The unit element is $(0,1)$ in $\{0\} \times {\mathbb Z}/5^*$, corresponding to $16$ in ${\mathbb Z}/20$.
This provides a way to construct many more examples. For example,
using ${\mathbb Z}/2 \times {\mathbb Z}/3 \cong {\mathbb Z}/6$ via $(x,y) \mapsto 3x - 2y$, the same
construction gives the multiplicative group $\{0\} \times {\mathbb Z}/3^*$.
The elements $(0,1)$ and $(0,2)$ of this group correspond to $4$ and $2$
of ${\mathbb Z}/6$.
A: Set $S=\{4,8,12,16\}$
The identity element of algebraic structure $(S,\times_{20})$ is $16$.



*

*Is this group closed under modulo multiplication: YES

*Is the operation associative: $(4\times_{20}8)\times_{20}12=12\times_{20}12=\textbf{4}=4\times_{20}(8\times_{20}12)=4\times_{20}16=\bf{4}$

*Is there exists identity : YES (16)

*Does each element has its own inverse : YES ($4 = 4^{-1}, 8 = 12^{-1}, 12= 8^{-1}, 16 = 16^{-1}$)


So, this algebraic structure is indeed a group under multiplication modulo 20.
A: If you want a simpler example look at $\{2,4\}$ modulo $6$; and a similar one $\{3,6,9,12\}$ modulo $15$. Also $\{7, 14\}$ or $\{3,6,9,12,15,18\}$ modulo $21$.
Take the last one, for example. Every element is divisible by $3$ but not by $7$. Their products will therefore be divisible by $3$ but not by $7$, and reducing $\bmod 21$ does not change divisibly by $3$ or $7$.
The set in fact comprises a complete set of non-zero residues modulo $7$. Since the products are all divisible by $3$, the value of he product modulo $21$ depends only on the value modulo $7$. The multiplicative identity is $15\equiv 1 \bmod 7$.
In your modulo $20$ example, if you reduce the elements modulo $5$ you get $\{4,3,2,1\}$ and the identity element gets the name you expect.
Note that the identity element of a group is defined by its properties, not its name.

You ask in your comments how your set is a subgroup - but you don't indicate what you think it is a subgroup of. But note that the non-zero integers modulo $20$ do not form a group under multiplication - we have $4\times 5=0$ for example, so the set is not closed under the proposed binary operation. You have to exclude multiples of $2$ and $5$ to get the standard group of invertible elements, but this does not contain the elements you are considering. No subgroup.

In the case where multiplication does not form a group we can have both non-trivial nilpotent $(e^2=0, e\neq 0)$ and idempotent $(e^2=e, e\neq 0,1)$ elements. Idempotents are candidates for the identity in groups made up of a subset of the original elements.
Modulo $20$ for example, $5$ is idempotent. Can you find a non-trivial multiplicative group modulo $20$ with $5$ as the identity?
