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This question already has an answer here:

Let $G = {5, 15, 25, 35}$. Prove that G is a group under multiplication modulo $40$. I know how to solve for associativity and closure but I don't know how to find the identity for the set. All the answers that I've seen are using the Cayley Table but I haven't learned that in school yet. Can someone please help me with another way to solve this? Any help will be appreciated.

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marked as duplicate by Chris Culter, Lord Shark the Unknown, Krish, Claude Leibovici, Patrick Stevens Sep 14 '17 at 7:16

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    $\begingroup$ Then learn it . $\endgroup$ – Kenny Lau Sep 14 '17 at 4:24
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    $\begingroup$ An identity solves $e^2=e$. Which of your elements solves that? $\endgroup$ – Lord Shark the Unknown Sep 14 '17 at 4:24
  • $\begingroup$ In making your multiplication table, you should write the entries first as the multiples of 25 that they are to simplify calculation slightly. After you solve this problem, it would be highly instructive for you to meditate on why this is a group, even though these 4 residues, and the number 40, all have a factor of 5 in them. What happens if you divide out the 5? What was the old group identity? Is it still the new one? Is there a natural correspondence between these two groups? $\endgroup$ – sigma-finite Sep 14 '17 at 4:33
  • $\begingroup$ See in particular the answer here: math.stackexchange.com/a/693704 and note the third duplicate question math.stackexchange.com/q/1114600 $\endgroup$ – Chris Culter Sep 14 '17 at 4:34
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Use the following result.(Don't just use, prove this!)

Consider the group $U(n)$ of units modulo $n$ under multiplication. For $m\in U(n)$ , $mU(n)$ is a group under multiplication mudulo $mn$. The identity elelment of $mU(n)$ is $m\beta $ where $\beta =m^{-1}$ in $U(n)$.

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  • $\begingroup$ "Just use it. (Don't just use it!)" $\endgroup$ – Patrick Stevens Sep 14 '17 at 7:16