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For $G, H$ given below, I want to build corresponding permutation groups. For $G, H$ given below, I want to build corresponding permutation groups.

$H: \langle\{2,4,8,10,14,16\}, \times_{18}\rangle$

$G: \langle \{3,6,9, 12,15,18\}, \times_{21}\rangle$

How do I achieve that?

Let the new groups be $G', H'$, but then what property is needed to find co-domain value for each element in them?

$G'$ : \begin{pmatrix}3&6&9&12 & 15& 18\\ &&&&&\end{pmatrix}

$H'$ : \begin{pmatrix}2&4&8&10& 14& 16\\ &&&&&\end{pmatrix}

This seems wrong, as $G', H'$ will be composed of some number of permutations.

If so, how many permutations are possible in either group $G', H'$?

$G: \langle \{3,6,9, 12,15,18\}, \times_{21}\rangle$

How do I achieve that?

Let the new groups be $G', H'$, but then what property is needed to find co-domain value for each element in them?

$G'$ : \begin{pmatrix}3&6&9&12 & 15& 18\\ &&&&&\end{pmatrix}

$H'$ : \begin{pmatrix}2&4&8&10& 14& 16\\ &&&&&\end{pmatrix}

This seems wrong, as $G', H'$ will be composed of some number of permutations.

If so, how many permutations are possible in either group $G', H'$?

jiten
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    This might be useful:$$a_{i} \rightleftharpoons P_{a_{i}}=\left(\begin{array}{cccc} a_{1} & a_{2} & \cdots & a_{n} \ a_{i} \cdot a_{1} & a_{i} \cdot a_{2} & \cdots & a_{i} \cdot a_{n} \end{array}\right)$$, where $a_i$ is an element of your $H$, and $P_{a_i}$ is corresponding permutation element. – narip Jul 28 '22 at 05:37
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    Hence $$a_{i} a_{j} \rightleftharpoons P_{a_{i} a_{j}}=\left(\begin{array}{cccc} a_{1} & a_{2} & \cdots & a_{n} \ a_{i} \cdot a_{j} \cdot a_{1} & a_{i} \cdot a_{j} \cdot a_{2} & \cdots & a_{i} \cdot a_{j} \cdot a_{n} \end{array}\right)$$ – narip Jul 28 '22 at 05:38
  • @narip Thanks, but as was a small group. So, if could give solution too. Or, need construct an even smaller group? – jiten Jul 28 '22 at 05:39
  • @narip Thanks, but don't where to look for it. The problem was not based on any source. So, if any search keywords or explicit term were given, please. – jiten Jul 28 '22 at 05:53
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    This may be helpful. There's a concrete example inside. – narip Jul 28 '22 at 06:08

2 Answers2

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For the permutation group, see this proof of Cayley's Theorem.

For example, the element of the permutation group corresponding to $3$ is

$$\begin{pmatrix} 3 & 6 & 9 & 12 & 15 & 18\\ 9 & 18 & 6 & 15 & 3 & 12 \end{pmatrix}.$$

This is because:

$$\begin{align} 3\times 3&=9,\\ 3\times 6&=18,\\ 3\times 9&\equiv 6\pmod{21},\\ 3\times 12&\equiv 15\pmod{21},\\ 3\times 15&\equiv 3\pmod{21},\\ 3\times 18&\equiv 12\pmod{21}. \end{align}$$

One can relable, so that the permutation corresponding to three is

$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 6 & 2 & 5 & 1 & 4 \end{pmatrix}.$$

Shaun
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  • Thanks, corrected the typo. of putting in $18$ in place of $3$ for $(3,6)$. Is it not just one permutation, or the group has this permutation (or some permutations of disjoint co-domains of elements) as the only one inside it? But, what will be the order of group then? – jiten Jul 28 '22 at 10:55
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    You're welcome, @jiten. If $K\cong L$, then $|K|=|L|$ for all groups $K, L$. – Shaun Jul 28 '22 at 11:00
  • To restate the last comment: there should be five more distinct permutations, as order of group $=6$. – jiten Jul 28 '22 at 11:00
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    Yes, @jiten. I gave an example only. I recommend that you compute the rest yourself. – Shaun Jul 28 '22 at 11:01
  • Relabel of the co-domain elements is the key then to create rest. But, that way get # of distinct permutations $=6!$, while should get only $6$. So, what limits the number of permutations? – jiten Jul 28 '22 at 11:06
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    No, @jiten. The number of permutations in the permutation group is six. Each of its elements corresponds to exactly one element of the original group and vice versa. See the proof of Cayley's Theorem. Try to understand my example above. Also, instead of writing another comment in five minutes, give it some time; be patient. – Shaun Jul 28 '22 at 11:10
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I thought that a different approach, combining existing answers, may help...

So consider your group that is multiplication modulo 18. You can work with powers of 2:

$H: \langle\{2,4,8,16,14,10\}, \times_{18}\rangle$ \begin{array}{c|cccccc}& 2&4& 8& 16&14& 10\\ \hline 2& 4& 8& 16& 14& 10& 2\\ 4& 8& 16& 14& 10& 2& 4\\ 8& 16& 14& 10& 2& 4& 8\\ 16& 14& 10& 2& 4& 8& 16\\ 14& 10& 2& 4& 8& 16& 14\\ 10& 2& 4& 8& 16& 14& 10\\ \end{array}

Note how the powers of $2$ increase as you go through the table to the right and down.

Then, according to @narip, each row can be thought of as a permutation... For example, look at the heading and the top row in the table. They are the permutation

$H'$ : \begin{pmatrix} 2& 4& 8& 16& 14& 10\\ 4& 8& 16& 14& 10& 2 \end{pmatrix}

The next row in the table corresponds to another permutation:

$H''$ : \begin{pmatrix} 2& 4& 8& 16& 14& 10\\ 8& 16& 14& 10& 2& 4 \end{pmatrix}

and so on...

The general form of the permutations is

$$H^a : \begin{pmatrix} 2^1& 2^2& 2^3& 2^4& 2^5& 2^6\\ 2^{1+a} & 2^{2+a} & 2^{3+a} & 2^{4+a} & 2^{5+a} & 2^{6+a} \end{pmatrix}(\bmod 18)$$

You can also take transposes of columns to arrive similarly at the result.

Matt Groff
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