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

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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.
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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.
 
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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.
 

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

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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