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I need little help for solving the following problem:

The value of $\alpha$ for which $G=\{\alpha,1,3,9,19,27\}$ is a cyclic group under multiplication modulo $56$ is which of the following?

(a)$5$,(b)$35$,(c)$25$,(d)$15$.

Can someone point me in the right direction? Thanks in advance for your time.

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2 Answers 2

up vote 2 down vote accepted

Well you know the group is cyclic, so how about you just compute powers of $3$?

$$ 1 \mapsto 3 \mapsto 9 \mapsto 27 \mapsto ? \mapsto ? $$ compute the last two $\pmod {56}$. Since $27 \cdot 3 \equiv 81 \equiv 25 \pmod{56}$, if this question makes any sense, you must have $\alpha = 25$ and $25 \cdot 3 \equiv 19 \pmod{56}$ and you know that $19 \cdot 3 \equiv 1 \pmod{56}$.

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$3^2=9, 3^3=27, 3^4=81=25$. So for your set to be a group at all, never mind a cyclic one, it has to contain $25$.

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thanks a lot sir.I have got it. –  user52976 Dec 16 '12 at 19:02
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