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1) Determine the Cayley Table of $(Z_5^*, \cdot)$

2) determine which additive group has the exact same table.

3) Further determine an isomorphism between those two groups and prove by means of that isomorphism that $(Z_5^*, \cdot)$ is a group

So since 5 is a prime number, each element different than 0 has a muliplicative inverse element and the Cayley table may be written as (is this correct?) $$\begin{array}{c|c|c|c|c} 5^{*} & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ \hline 2 & 2 & 4 & 1 & 3 \\ \hline 3 & 3 & 1 & 4 & 2 \\ \hline 4 & 4 & 3 & 2 & 1 \end{array}$$

So we see that [4] is its own multiplicative inverse and [2] and [3] are the multiplicative inverses of each other. However I have no idea what to do about questions 2) and 3).

M-xyz1
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    Your table looks correct. It is isomorphic to $\mathbb Z/4\mathbb Z$ with $+$. There are only two groups of order $4$, up to isomorphism, and only one of them is cyclic – J. W. Tanner Jun 08 '20 at 01:29
  • @J.W.Tanner. Thank you, $\mathbb{Z}/4\mathbb{Z}$ is this one, right? https://proofwiki.org/wiki/Groups_of_Order_4 (groups are a completely new topic for me), however I can not find a one-to-one mapping of the elements for these two groups? – M-xyz1 Jun 08 '20 at 01:55
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    Right. $(\mathbb Z/5\mathbb Z)^\times, \times$ and $\mathbb Z/4\mathbb Z, +$ are both cyclic; map a generator to a generator (an element of order $4$ to an element of order $4$) – J. W. Tanner Jun 08 '20 at 02:01
  • @J.W.Tanner, thank you for your helpful comments, however I am still having a little bit difficulties with the mapping, I think the element 0 from $\mathbb{Z}/\mathbb{4Z}$ should be mapped to the element 4 from the other group, but what to do with the other ones? – M-xyz1 Jun 08 '20 at 02:25
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    The element $0$ in $\mathbb Z/4\mathbb Z,+$ is the identity, so should be mapped to the identity $1$ in $(\mathbb Z/5\mathbb Z)^\times, \times$. Can you identify an element of order $2$ in each group? – J. W. Tanner Jun 08 '20 at 02:44
  • @J.W.Tanner: In $(\mathbb{Z}/5\mathbb{Z})^{\times}$ $4$ is of order 2, and in $(\mathbb{Z}/4\mathbb{Z})$ $2$ is of order 2. And $3$ is in both groups of order 4. in $(\mathbb{Z}/4\mathbb{Z})$ $1$ is of order 4, and in $(\mathbb{Z}/5\mathbb{Z})^{\times} $ $2$ is of order 4. Is it correct until now? – M-xyz1 Jun 08 '20 at 11:21
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    Yes, that is correct. And isomorphisms preserve the orders of the group elements – J. W. Tanner Jun 08 '20 at 13:53
  • @J.W.Tanner: thank you so much for taking the time to check my answer!, I have another question, for example in each group there are two elements of order $4$, can I choose which will be mapped to which? and how can I write the isomorphism $\phi: (\mathbb{Z} / 5 \mathbb{Z} \rightarrow (\mathbb{Z} / 4 \mathbb{Z}, +)$ mathematically precise? and last, how can I use the isomorphism to prove that $\mathbb{Z} / 5 \mathbb{Z}$ is indeed a group? Does this stem from the fact that this isomorphism exists? – M-xyz1 Jun 08 '20 at 15:49

2 Answers2

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Note that $1 = 2^4, 2 = 2^1, 3 = 2^3, 4 = 2^2$.

So there is an isomorphism from $\mathbb Z_4$ under addition to $\mathbb Z_5$ under multiplication, taking $n$ to $2^n$.

hdighfan
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$\mathbb Z/4\mathbb Z, +$ is isomorphic to $(\mathbb Z/5\mathbb Z)^\times,\times$.

Both have two elements of order $4$: $1,3\in\mathbb Z/4\mathbb Z$ and $2,3\in(\mathbb Z/5\mathbb Z)^\times$.

An isomorphism will map one of these elements in $\mathbb Z/4\mathbb Z$ to one of those elements in $(\mathbb Z/5\mathbb Z)^\times$ .

So there are actually two isomorphisms from $\mathbb Z/4\mathbb Z$ to $(\mathbb Z/5\mathbb Z)^\times$ :

one maps $1\mapsto2$, and the other maps $1\mapsto3$.

Since $1$ generates the elements of $\mathbb Z/4\mathbb Z$, once the image of $1$ is determined,

the images of all the elements are determined; e.g., $\phi(3)=\phi(1+1+1)=\phi(1)^3$.

Determining the isomorphism in this way ensures that $(\mathbb Z/5\mathbb Z)^\times,\times$ is a group, just like $\mathbb Z/4\mathbb Z, +$.

To write the isomorphisms the other way around, $\Psi(2)=1$ or $3$ and $\Psi(2^n)=n\Psi(2)$.

J. W. Tanner
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  • Thank you so much for the detailed answer, the things which I am not sure I understood correctly are 1) $\Phi(2^n)$ I have to regard the $2^n$ mod(5), right? so for example if I want to get the element $3$ I have to set n=3, since $2^3=8=3(5)$, for $1$, n=4, and so on, correct? and how do I know that $\phi(1+1+1)={\phi(1)}^{3}$, why can I write the sum as power? (I mean if I take the numbers, it is clear why, but am I allowed to do this always? – M-xyz1 Jun 09 '20 at 01:19
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    Yes, regard as modular arithmetic; and to make it an isomorphism, addition in $\mathbb Z/4\mathbb Z$ corresponds to multiplication in $(\mathbb Z/5\mathbb Z)^\times$ – J. W. Tanner Jun 09 '20 at 01:28
  • Thank you and could you please also tell me how one can yield $\Phi(2^n)=n\Phi(2)$ is it just by trying or is there a structure when doing this? – M-xyz1 Jun 09 '20 at 01:47
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    Since $\Psi$ is an isomorphism, $\Psi(2\times...\times2)=\Psi(2)+\cdots+\Psi(2)$ – J. W. Tanner Jun 09 '20 at 01:50
  • Thank you again, for the explanations – M-xyz1 Jun 10 '20 at 00:10
  • @J.W.Tanner Please help as have some more doubts still. (1) To ensure Isomorphism between the two groups, is it needed that the multiplicative group is of prime order? (2) Have learned that a multiplicative operation is implemented by addition operation, and power table by multiplication operation. But, that should mean: $\left<(\mathbb Z/n)^\times,\times\right>$ implements a power table. And $\left<\mathbb Z/(n-1)\mathbb Z, +\right>$ implements a multiplication table? – jiten Oct 26 '22 at 06:39