# Proving G is Cyclic & G=<a,b> with #G=77

• nowits
In summary, in this conversation about a group G with order 77, the main focus is on proving two statements. One is that if there exists an element a in G that satisfies a21≠1 and a22≠1, then G is cyclic. The other is that if there are elements a and b in G with orders 7 and 11 respectively, then G is generated by a and b. It is also mentioned that any proper subgroups of G must have orders 7 or 11. The approach for (a) is to use the fact that 21= 3*7 and 22= 2*11, while for (b) it is noted that this statement is trivial.
nowits

## Homework Statement

Let G be a group and let #G=77. Prove the following:
a) G is cyclic, if there is such an element a in G that a21≠1 and a22≠1
b) If there are such elements a and b, so that ord(a)=7 and ord(b)=11, then G=<a,b>

2. Homework Equations , 3. The Attempt at a Solution
I really don't even know where to begin with these. So I'd appreciate if someone could point me in the right direction.

You do understand, don't you, that any proper subgroups must be of order 7 and 11? And that are subgroups of those orders? That should make (b) trivial.

As for (a) the crucial point is that 21= 3*7 and 22= 2*11.

## 1. How do you prove that a group G is cyclic?

To prove that a group G is cyclic, you need to show that there exists an element in G that can generate all other elements in the group. This element is called a generator and if it exists, then G is considered to be a cyclic group.

## 2. What is the significance of #G=77 in proving that G is cyclic?

The notation #G=77 means that the group G has 77 elements. This is important because it helps determine the order of the group, which is the number of elements in the group. In order for G to be cyclic, the order of the group must be finite.

## 3. How do you show that G= with #G=77?

To show that G= with #G=77, you need to prove that a and b are generators of G and that they have an order that is a factor of 77. This means that the smallest positive integer k such that a^k = e (identity element) and b^k = e must divide 77.

## 4. Can G be cyclic if #G=77 but G is not equal to ?

Yes, it is possible for G to be cyclic even if it is not equal to . In order for G to be cyclic, it must have at least one generator. As long as there exists an element in G that can generate all other elements, then G can be considered cyclic.

## 5. What is the relationship between proving G is cyclic and showing that G is isomorphic to another group?

If you can prove that a group G is cyclic, then it automatically means that G is isomorphic to another group. This is because cyclic groups have a very specific structure that can be mapped to other groups through an isomorphism. Therefore, proving that G is cyclic can also be seen as showing that G is isomorphic to a cyclic group.

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