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This question already has an answer here:

Show that the set $\{5, 15, 25, 35\}$ is a group under multiplication modulo $40$. Is there a relation between this and $U(8)$?

I am having a really difficult time beginning this proof. All this is abstract to me.

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marked as duplicate by Chris Culter, Lord Shark the Unknown, Krish, user296602, Jeremy Rickard Sep 14 '17 at 15:35

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Hint: Since there are only four elements (and the group is abelian because multiplication is commutative), it's easy to write down the multiplication table and check the axioms directly.

For example, one finds that $25 \cdot g = g$ for all $g$ in the set, so $25$ is the identity.

For the associativity axiom, one need not use the multiplication table, as we already know the group operation (multiplication modulo a given number) is associative.

There is a relationship between this and the ring $U(8)$. Namely, we can consider the group $U(8)^* = \{1, 3, 5, 7\}$ of invertible elements in $U(8)$, and this turns out to be isomorphic to the given group (both are isomorphic to the Klein $4$-group, $U(2) \times U(2)$).

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  • $\begingroup$ Thank you. I am not familiar with the definitions of the similarity between them but I could follow the proof. $\endgroup$ – Karina Pena Jan 22 '15 at 4:46
  • $\begingroup$ The similarity here (isomorphism) can be thought of as saying that the multiplication tables of the two groups are the same, simply with the elements renamed. $\endgroup$ – Travis Jan 22 '15 at 5:50
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Well, it is abstract algebra... Jokes aside, do you know the axioms of a group? You need to check, for each pair of elements in the set, that the product mod $40$ is still in the set, find the identity (should surface when you check for closure), and an inverse for each element. For a set like this one, you can do each calculation by hand (there are after all just three possibilities for each inverse). The associativity law follows from the associativity of integers.

Once you have that it is a group, you can set up the multiplication table, and compare it to the multiplication table of $U_8$. See if you can spot any similarities.

Comment if you need me to fill in more details!

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  • $\begingroup$ This is my first course in algebra and although I understand the reading when it's time for me to do the problems I am at a halt. Any pointers? $\endgroup$ – Karina Pena Jan 22 '15 at 4:43
  • $\begingroup$ I do not know how to begin the proof, I may be over thinking it because reading these answers makes sense. $\endgroup$ – Karina Pena Jan 22 '15 at 4:44
  • $\begingroup$ @KarinaPena It takes a while to get used to the kind of thinking required. The first step is to verify that it is closed under the group operation. You need to show that the product of every pair of elements is a group element. This is done by just trying every pair. Can you do that? $\endgroup$ – Johanna Jan 22 '15 at 4:46
  • $\begingroup$ Just show that 5•15, 5•25, and 5•35 is in the set? And follow the same with the rest of the elements? $\endgroup$ – Karina Pena Jan 22 '15 at 4:51
  • $\begingroup$ @KarinaPena No, that the elements $5*5, 5*15, 5*25, 5*35, 15*25, \dots, \pmod{40}$ are in the set. $\endgroup$ – Johanna Jan 22 '15 at 4:54

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