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Im working on a problem for my class and am stuck on the last part of it, any help is appreciated.

Let $G = U_{19}$ the units of $Z/19Z$. Find the cyclic subgroups generated by $<7>, <12>, <8>$

I found that $<7> = [7,11,1]$

For $<12>$ I noticed that $ 12 = -7 \mod19$ thus $<12>$ must have order 6 and has elements $[7,11,-1,-7,-11,1]$ by inspecting powers of $-7$.

Now for $<8>$, I notice that $7^2 = -8 = 11$. I also see that $-8$ then should be $-7^5 = 12^5$ and im confident this contains the key to the problem, but ive been staring at this for 45 minutes now and cant seem to make the jump to how to use this information to figure out the elements of $<8>$. Am I missing a piece of information or am i just not getting it? Thanks again!

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1  
8 * 8 = 64 == 7. 8 * 7 = 56 == 18. 8 * 18 = 144 == 11. 8 * 11 = 88 == 12. 8 * 12 = 96 == 1. 8 * 1 = 8 == 8. So you get <1,7,8,11,12,18>. –  Will Jagy Nov 4 '12 at 20:40

1 Answer 1

up vote 1 down vote accepted

Just compute $g,g^2,g^3,g^4,\ldots$ until you get the identity:

7, 11, 1
12, 11, 18, 7, 8, 1
8, 7, 18, 11, 12, 1

so we have

  • $\langle 7 \rangle = \{1,7,11\}$
  • $\langle 12 \rangle = \{1,12,11,18,7,8\}$
  • $\langle 8 \rangle = \{1,8,7,18,11,12\}$

I'll show the full details of the third calculation, everything is done mod 19:

8
8*8  = 64  = 7
8*7  = 56  = 18
8*18 = 144 = 11
8*11 = 88  = 12
8*12 = 96  = 1, done
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I see that you calculated <8> directly, but isnt there something to notice about how <8> is the same as a power of <12>? I notice that the elements of <8> and <12> are the same- surely this is no coincidence and im guessing is a result of the fact that 8 = 12^5 –  MSEoris Nov 4 '12 at 21:49
    
@MSEoris, yeah you're right $\langle 8 \rangle = \langle 12 \rangle$. –  sperners lemma Nov 4 '12 at 21:54
1  
@MSEoris, in fact $8^{-1} \equiv 12 \pmod {19}$ –  sperners lemma Nov 5 '12 at 12:46

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