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U(49) under multiplication modulo 49 ...
Just wondering how the heck they get 42 elements, are these elements supposed to be relatively prime to 49
7, 14, 21, 28, 35, 42, 49 are not relatively prime to 49; if you take away these 7 numbers, you have 42 numbers left. i think this is why U(49) has 42 elements.
but forget about why. they told u that the group is cyclic and has 42 elements. so just use this fact
"If d is a positive divisor of n, the number of elements of order d in a cyclic group of order n is phi(d)"
restate the theorem: Let G be a cyclic group of order n. If d | n, then the number of elements of order d in G is phi(d).
use the theorem: G = U(49) has order n = 42.
we want to know how many generators in G (that is, how many elements of order n; recall that g generates G if G = )
we have n | n. so by theorem, the number of elements of order n in G is phi(n).
so the number of generators in G is phi(n) = phi(42) = ....
_________________ The word analysis comes from the Greek ανάλυσις, which means "dissolving into pieces."http://www.facebook.com/home.php?ref=ho ... 2306439078
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