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

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.

• 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. – Hans Engler May 27 at 2:14
• 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. – N_Mathematics_B May 27 at 3:56

A group G is cyclic if there is an element g $$\in$$ G such that 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=$$ by $$f(n)=ng$$. Note that you must prove that this map is a bijective homomorphism.

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.

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).