Abstract Algebra help

[From: ] [author: ] [Date: 11-12-12] [Hit: ]

. (since g^2 = e)= gh^2g= g(ghg)g= g^2hg^2= ehe= h===> h^3 = eThus h is of order 3.-To fill in a gap:2) U(p) stands for the multiplicative group of (nonzero) elements mod p, with p prime.In general, U(p) is cyclic of order p-1.......

4. ga = g ===> g^-1ga = g^-1g ===> ea = e ===> a = e

But, ga = g holds for all a in G, so a = e holds for all a in G. That is, G = {e}.

5. If ghg = h^2, then:

h^4 = (ghg)^2 = ghg^2hg
= ghehg ... (since g^2 = e)
= gh^2g
= g(ghg)g
= g^2hg^2
= ehe
= h
===> h^3 = e

Thus h is of order 3.

-

To fill in a gap:

2) U(p) stands for the multiplicative group of (nonzero) elements mod p, with p prime.
In general, U(p) is cyclic of order p-1.

For p = 11:
One such generator is 2, since 2^5 = -1 (mod 11) while 2^10 = (-1)^2 = 1 (mod 11).
Since |U(11)| = 10, the generators of U(11) are of the form 2^k, where gcd(k, 10) = 1.
==> The generators are 2^1 = 2, 2^3 = 8, 2^7 = 7, 2^9 = 6 (mod 11).

As for an isomorphism, we only need to map a generator of U(11) to a generator of Z10:
Let f : U(11) --> Z10 via f(2) = 1 (or more generally f(2^k) = k).
Since every element in U(11) can be written in the form 2^k, f is defined on U(11).
That this is a homomorphism and bijective is easy to verify.

I hope this helps!

12

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