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For each of the groups $\mathbb Z_4$,$\mathbb Z_4^*$ indicate which are cyclic. For those that are cyclic list all the generators.

Solution

$\mathbb Z_4=${0,1,2,3}

$\mathbb Z_4$ is cyclic and all the generators of $\mathbb Z_4=${1,3}

Now if we consider $\mathbb Z_4^*$

$\mathbb Z_4^*$={1,3}

How do i know that $\mathbb Z_4^*$ is cyclic?

In our lecture notes it says that the $\mathbb Z_4^*$ is cyclic and the generators of $\mathbb Z_4^*$=3

Can anyone help me on the steps to follow in order to prove the above?

M.Sina
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Timoci Lagilevu
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    Perhaps you should start by writing out the group multiplication table for $\mathbb Z^_4$. But one way to know the answer is that $\mathbb Z^_4$ has only two elements; there is only one group of order 2, and so $\mathbb Z^*_4$ must be isomorphic to $\mathbb Z_2$. – MJD Feb 21 '13 at 03:33
  • Step 1: make sure you know the definitions of cyclic and generator. Step 2: apply what you know to that group and that generator. – Gerry Myerson Feb 21 '13 at 03:33
  • Can a (the) group with two elements not be cyclic? – Julien Feb 21 '13 at 03:35
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    may help: http://math.stackexchange.com/questions/290427/is-mathbb-z-p-1-2-3-p-1-a-cyclic-group –  Feb 21 '13 at 03:37
  • @MJD in that case if we consider $\mathbb Z_10^*$ it has 4 elements namely 1,3,7,9. Does it mean that it is isomorphic to $\mathbb Z_4$ – Timoci Lagilevu Feb 21 '13 at 03:49
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    No, there are two different groups with 4 elements, and only one of them is $\mathbb Z_4$. The trick only works when the number of elements is a prime. – MJD Feb 21 '13 at 03:54
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    In addition to MJD, after you write out the Cayley Table for $\mathbb Z_4^$, the table is supposed to contain only the two elements in $\mathbb Z_4^$ which are 1,3. By using the same way you found the generators for $\mathbb Z_4$, do so for finding the generators in $\mathbb Z_4^$. You will then see that the only generator is 3, as 1 is the identity element. So then, it is true that $\mathbb Z_4^$ is cyclic and the only generator is 3. – Faye Feb 21 '13 at 20:13

2 Answers2

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Its a group with two elements and therefore must be isomorphic to $\mathbb{Z}_2$ a simple check will show you which one is not the identity element and therefore the generator of the group.

Sean Ballentine
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Consulting here, you can easy see that in modulo $4$ there are two relatively prime congruence classes, $1$ and $3$, so $(\mathbb{Z}/4\mathbb{Z})^\times \cong \mathrm{C}_2$, the cyclic group with two elements.

Mikasa
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