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.

  • 1
    $\begingroup$ Funny, I thought the identity element of a group was something satisfying $e*g=g*e=g$ for all $g$, not something with the label "$1$". $\endgroup$ – Lord Shark the Unknown Jun 13 '19 at 5:26
  • $\begingroup$ Build a cayley table? $\endgroup$ – Vineet Jun 13 '19 at 5:29
  • $\begingroup$ I have built Cayley table. $\endgroup$ – user679537 Jun 13 '19 at 5:48
  • 1
    $\begingroup$ It appears the identity element is 16 since $16×_{20}g=g×_{20}16=g$. One part of the definition is identity: There exists an element $e$ such that $e×g=g×e=g$ for all $g\in G$. Well there it is... $e=16$. The number $1$ is not in the set so can't be considered. $\endgroup$ – Pixel Jun 13 '19 at 6:12

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 ${\mathbb Z}/6$.


Set $S=\{4,8,12,16\}$

The identity element of algebraic structure $(S,\times_{20})$ is $16$.

enter image description here

  • 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.

  • 1
    $\begingroup$ This does not address the OP’s doubt. $\endgroup$ – Carsten S Jun 13 '19 at 5:56
  • $\begingroup$ But how is it possible that 16 is idenity element with respect to multiplication modulo 20 because 1 is an idenity w r t the same operation. $\endgroup$ – user679537 Jun 13 '19 at 5:57
  • $\begingroup$ @CarstenS Yes, it does not.. But I think OP somehow got confused with $1$ as identity. $\endgroup$ – Vineet Jun 13 '19 at 6:00
  • $\begingroup$ Not actually,I am the OP.I know the identity element is 16.But it is somewhat strange because the operation multipliction modulo 20 has idenity 1 when operated on U(20) but has identity 16 when operated on {4,8,12,16}.It mean that identity depends not only on the operation but also on the underlying set. $\endgroup$ – user679537 Jun 13 '19 at 6:03
  • 1
    $\begingroup$ You can mail me on kishalaysarkar2000@gmail.com whenever you find any new detail on this question or want to discuss on this question. $\endgroup$ – user679537 Jun 13 '19 at 6:54

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?

  • $\begingroup$ I did not understand the last paragraph of your answer. $\endgroup$ – user679537 Jun 13 '19 at 6:12
  • $\begingroup$ @KishalaySarkar (With the operation of multiplication, the non-zero multiples of $4$ are not a subgroup of the non-zero residues modulo $20$. One reason is because the non-zero residues modulo $20$ do not form a group with this operation (if it were a group it would have $19$ elements and no non-trivial subgroups). I've added a little more o the answer. $\endgroup$ – Mark Bennet Jun 13 '19 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy