- #1
CAF123
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Homework Statement
1)Fix ##n \in \mathbb{N}##. Consider multiplication mod ##n##. Let G be the subset of {1,2,...,n-1} = ##\mathbb{Z}_n## \ ##\left\{0\right\}## consisting of all those elements that have a multiplicative inverse (under multiplication mod n). Show that G is a group under multiplication. Describe the group when n =12.
2) Show that if G is cyclic, then G is abelian.
The Attempt at a Solution
1) The reason I can't make much progress here is essentially interpretation. '..that have a multiplicative inverse (under multiplication mod n)' What does this mean?
What I said intially was multiply a number, p, in {1,2...n-1} by x and then take that number mod n and if there exists an x such that this equals 1, then that is in G. However, considering some cases, n=3,4 etc.. there exists an x for all elements. I tried something else that made me get a single element for G for all n. What is the appriopriate interpretation?
2)I know the proof to this since it is in my textbook, but I tried it slightly different and I want to know if it is valid:
Consider two elements, ##g_1, g_2##, with ##g_1## the generator of the group and ##g_2## some element generated by ##g_1##. Then consider the case of ##g_1 = g_2##. Since G is cyclic, this means, ##g_2 = g_1^n = g_1##, n the order of ##g_1##.
Now multiply by ##g_2## on both sides: ##g_1 g_2 = g_2 g_2 = g_2 g_1^n = g_2 g_1##.
Many thanks.