1. ## Is My Answer Close to Right? Cyclic Group Proof

Suppose that G is a group of order 143 and that H is a subgroup of G with H not equal to G. Prove H is cyclic.

Solution Attempt:
Suppose G is a group and |G| = 143.
Thus, G has an element a with order 13 and an element b with order 11. Then ab has order 143 since

(ab)^143 = (a^143)(b^143) = e. However, (ab)^13= (a^13)(b^13) = b^13 not equal to e.
Also, (ab)^11 = (a^11)(b^11) = a^11 not equal to e. Then G = <ab> is the cyclic group generated by ab.

A Theorem says that any subgroup of a cyclic group is cyclic, thus H is cyclic because G is cyclic and H is a subgroup of G.

Thanks! Any feedback would be most most appreciated.

2. Originally Posted by crystalwater
Suppose that G is a group of order 143 and that H is a subgroup of G with H not equal to G. Prove H is cyclic.

Solution Attempt:
Suppose G is a group and |G| = 143.
Thus, G has an element a with order 13 and an element b with order 11. Then ab has order 143 since

(ab)^143 = (a^143)(b^143) = e. However, (ab)^13= (a^13)(b^13) = b^13 not equal to e.
Also, (ab)^11 = (a^11)(b^11) = a^11 not equal to e. Then G = <ab> is the cyclic group generated by ab.

A Theorem says that any subgroup of a cyclic group is cyclic, thus H is cyclic because G is cyclic and H is a subgroup of G.

Thanks! Any feedback would be most most appreciated.
Not quite...in your proof that your element $\displaystyle ab$ has order 143 you assume that your group is abelian. This is not necessarily the case.

Essentially, you are looking at groups of order $\displaystyle pq$ with $\displaystyle p$ and $\displaystyle q$ primes. These are not all abelian - take $\displaystyle S_3$ for example.

To prove your question, what can the order of your subgroup $\displaystyle H$ be?

3. Originally Posted by crystalwater
Suppose that G is a group of order 143 and that H is a subgroup of G with H not equal to G. Prove H is cyclic.

Solution Attempt:
Suppose G is a group and |G| = 143.
Thus, G has an element a with order 13 and an element b with order 11. Then ab has order 143 since

(ab)^143 = (a^143)(b^143) = e. However, (ab)^13= (a^13)(b^13) = b^13 not equal to e.
Also, (ab)^11 = (a^11)(b^11) = a^11 not equal to e. Then G = <ab> is the cyclic group generated by ab.

A Theorem says that any subgroup of a cyclic group is cyclic, thus H is cyclic because G is cyclic and H is a subgroup of G.

Thanks! Any feedback would be most most appreciated.

In order to conclude that $\displaystyle (ab)^{143}=a^{143}b^{143}$ you must be assuming that either $\displaystyle G$ is abelian or else, at least, that $\displaystyle a,b$ commute. This is the case in fact, since there's only one group of order 143 up to isomorphism, the cyclic one...but I can't see how to prove this without using Sylow theorems, which you didn't even bring up.
The continuation is right.

Tonio

4. The Lagerange theorem says that the order of subgroup is the divisor of the oder of the group.
What is the possible value of the oder of a subgroup in G?