# 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.

• 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 "$1$". – Lord Shark the Unknown Jun 13 '19 at 5:26
• Build a cayley table? – Vineet Jun 13 '19 at 5:29
• I have built Cayley table. – user679537 Jun 13 '19 at 5:48
• 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. – 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 $$16$$. • 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.

• This does not address the OP’s doubt. – Carsten S Jun 13 '19 at 5:56
• 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. – user679537 Jun 13 '19 at 5:57
• @CarstenS Yes, it does not.. But I think OP somehow got confused with $1$ as identity. – Vineet Jun 13 '19 at 6:00
• 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. – user679537 Jun 13 '19 at 6:03
• You can mail me on kishalaysarkar2000@gmail.com whenever you find any new detail on this question or want to discuss on this question. – 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?

• I did not understand the last paragraph of your answer. – user679537 Jun 13 '19 at 6:12
• @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. – Mark Bennet Jun 13 '19 at 6:42