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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.

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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    $\begingroup$ Try not to get confused thinking about the multiplicative group of units of $\Bbb Z_{27}$ as the additive cyclic group $\Bbb Z_{18}$ :) $\endgroup$ Commented Jun 20, 2013 at 14:39
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    $\begingroup$ I'm not sure why the order of $2$ in $\mathbb Z_{18}$ tells you anything about the order of $2$ in $\mathbb Z_{27}^\times$. Yes, these two groups are isomoporphic, but the $2$ in one does not correspond to the $2$ in the other. $\endgroup$ Commented Jun 20, 2013 at 14:41
  • $\begingroup$ @rschwieb Are they different? Since they are 'finite abelian', I have thought that they can be isomorphic... I'm very confused... 'abstract algebra' is too 'abstract' :( $\endgroup$ Commented Jun 20, 2013 at 14:51
  • $\begingroup$ @user73309 No, what you wrote is fine! I just didn't want the fact that one was written multiplicatively and the other is written additively to confuse you. There isn't anything wrong with what you said. $\endgroup$ Commented Jun 20, 2013 at 16:04
  • $\begingroup$ 2 has multiplicative order 18 in $\Bbb Z_{27}$ as you noted below. When it is mapped into $\Bbb Z_{18}$ however, it is mapped to one of the generators of $\Bbb Z_{18}$: $\{1,5,7,11,13,17\}$. $\endgroup$ Commented Jun 21, 2013 at 11:44

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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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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.

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  • $\begingroup$ Finding the map between them could be the real problem. Thanks! $\endgroup$ Commented Jun 20, 2013 at 15:00
  • $\begingroup$ @user73309 Well, fortunately there are only finitely many ways to map :) $\endgroup$ Commented Jun 20, 2013 at 16:04
  • $\begingroup$ '2' in $\mathbb{Z}_{27}$ is of order 18 and it generates all elements which are relatively prime to 27. Then, could I say '2' is 'the' generator of the unit group of $\mathbb{Z}_{27}$? However it should have 'two' generators from my attempt....????? $\endgroup$ Commented Jun 20, 2013 at 21:34
  • $\begingroup$ @user73309 If you want to go about it this way, then you don't need to find elements in $\Bbb Z_2\times \Bbb Z_9$. From the 18 elements which 2 generates multiplicatively, you can just take the ones which are powers of 2 which are coprime to 18. $\endgroup$ Commented Jun 21, 2013 at 10:25
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As you note $|\mathbb{Z}_{27}^{\times}|=18$ and the straightforward calculations shows that $|\bar 2|=18$ so $\mathbb{Z}_{27}^{\times}=\langle \bar 2 \rangle$. To find all generators, consider the following:

$|2^k|=\frac{18}{(k,18)}=18\iff k=1,5,7,11,13,17$.

Hence all generators are $\bar 2, \overline{2^5},\overline{2^7},\overline{2^{11}},\overline{2^{13}},\overline{2^{17}}$.

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