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Let A = {1, 3, 5, 9, 11, 13} and let $\odot$ define the binary operation of multiplication modulo 14.

  1. Prove that (A, $\odot$ ) is a group. (You may assume that multiplication is associative.)

  2. List the cyclic subgroups of (A, $\odot$ ).

  3. Explain why (A, $\odot$) is not isomorphic to the symmetric group $S_3$.

  4. State an isomorphism between (A, $\odot$) and ($\mathbb{Z}_6$, +).

So, I understand that in part (1) to prove that it is, in fact, a group I must show that it is associative, has closure, has an identity element, as well as inverses. In part (2) I get confused, I don't quite understand what a cyclic group is, and I have no idea how to find them to list them.

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    $\begingroup$ Hello and welcome to math.stackexchange. To approach 1, 3, and 4, make multiplication tables of the three groups that are involved here and compare. $\endgroup$ – Hans Engler May 27 at 2:14
  • $\begingroup$ Quick question. While doing part (1) I was able to show associativity, closure, and identity of the set. However, I am unsure how to show that the set has inverses since my Caylee table contradicts it I think. $\endgroup$ – N_Mathematics_B May 27 at 3:56
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A group G is cyclic if there is an element g $\in$ G such that G=$<g>$. You can show that your group A is in fact cyclic and that is because there is an element of order 6. Since A is cyclic, a subgroup H of G will also be cyclic and therefore will be generated by one of the elements in G. To show A is not isomorphic to the symmetric group $S_3$ it suffices to note that A is abelian while $S_3$ is not. Lastly, an isomorphism between two groups is a bijective homomorphism. Because A is cyclic, and finite it must be isomorphic to $\mathbb{Z}_n$ for some integer n. You can define a map from $\mathbb{Z}_n$ to a cyclic group $G=<g>$ by $f(n)=ng$. Note that you must prove that this map is a bijective homomorphism.

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A cyclic group is one that is generated by one element and its inverse. For finite groups, a cyclic group is isomorphic to the addition modulo the number of elements. The isomorphism you are to show in part $4$ shows the whole group is cyclic. All subgroups of a cyclic group are also cyclic, but you should list them and show the generator for each.

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Write out the $6$ by $6$ multiplication table to verify that $1$ is the identity element, each element has an inverse, and closure is satisfied.

Computing the powers of $3$, and one obtains:

$3^1=3$

$3^2=9$

$3^3= 13$ (because $27$ mod $14$ = $13$)

$3^4 = 11$ (because $3^4 = 3^3 \cdot 3 = 13 \cdot 3 = 39 = 11$)

$3^5 = 5$

$3^6 = 1$.

Thus, the powers of $3$ generate the entire group, whence the group is cyclic.

The map $3^i \rightarrow i$ from the powers of the generator of the group to the powers of the generator of $\mathbb{Z}_6$ gives an isomorphism from the group to $\mathbb{Z}_6$.

The cyclic subgroups of $\mathbb{Z}_n$ are exactly the cyclic groups generated by $d$ for each divisor $d$ of $n$. The cyclic subgroups of $A$ are therefore $\langle 3 \rangle$, $\langle 3^2 \rangle$, $\langle 3^3 \rangle$, and $\langle 3^6 \rangle$.

The given group and $S_3$ are not isomorphic because a cyclic group is abelian and $S_3$ is non-abelian (or, you can say that $S_3$ is non-cyclic whereas the given group is cyclic).

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