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“Find all generators of $ (\mathbb{Z}_{27})^{\times} $”

My attempt is below.

Since $ (\mathbb{Z}_{n})^{\times} $ is a cyclic if and only if $ n = 1, 2, 4, p^n, 2p^n $, $ (\mathbb{Z}_{27})^{\times} $ is cyclic.

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And the order of $ (\mathbb{Z}_{27})^{\times} $ is $ 3^3 – 3^2 = 18 $ by the Euler’s phi function.

Thus $ (\mathbb{Z}_{27})^{\times} \cong \mathbb{Z}_{18} \cong \mathbb{Z}_2\times\mathbb{Z}_9 $.

But I don’t know how to get the ‘all’ generators.

(‘2’ seems to be the ‘one’ generator, since 2 is of order 9 in $\mathbb{Z}_{18}$.)

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Hint: If you are viewing the group as $\Bbb Z_2\times\Bbb Z_9$, you are just looking for elements (pairs!) which have additive order 18. The orders of elements of the form $(a,0)$ and $(0,b)$ are all clear to you… but do you realize what the orders of elements of the form $(a,b)$ are?

If you really intend to chase these elements back to $(\Bbb Z_{27})^\times

$, you will have to explicitly write the map you have between this and $\Bbb Z_2\times\Bbb Z_9$.

With lhf’s hint, this is the way to go.

If you know one generator $g$ of a cyclic group of order $n$, then all others are of the form $g^k$ with $(k,n)=1$.

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