I need some serious help with these questions. I have no clue where to start:

(G denote a group)

1. Let G be cyclic of order 36 and a be in G with a^12 not equal to, e (e is not equal to both of them), and not equal a^18. Prove G = <a>.

2. Let G be abelian of order p*g, where p and g are different primes. Show G is cyclic.

3. Let sigma be the following permutation in S_6: sigma = (134)(1562). Find sigma^38. (Write as a permutation not a product of cycles.)

4. Suppose k is a proper subgroup of H and H is a proper subgroup of G. If K=|420| and G=|42| what are th possible orders of H? Provide Arguments.

Any help and all hints are appreciated. Thanks guys!!!

2. Originally Posted by Luck of the Irish
1. Let G be cyclic of order 36 and a be in G with a^12 not equal to, e (e is not equal to both of them), and not equal a^18. Prove G = <a>.
What is $\text{ord}(a)$. It needs to divide $36$ those conditions force the order to be just that.

2. Let G be abelian of order p*g, where p and g are different primes. Show G is cyclic.
If you find an element $a$ of order $p$ and an element $b$ of order $g$ then order of $ab$ is $pg$.

3. Originally Posted by Luck of the Irish
3. Let sigma be the following permutation in S_6: sigma = (134)(1562). Find sigma^38. (Write as a permutation not a product of cycles.)
Note that $(134)$ and $(1562)$ commute. Therefore $[(134)(1562)]^{34} = (134)^{34}(1564)^{34} = (134)(1564)^2$
Can you simplify that?

4. Suppose k is a proper subgroup of H and H is a proper subgroup of G. If G=|420| and H=|42| what are th possible orders of H? Provide Arguments.
Remember $|H|$ divides $|G| = 420$ and $42=|K|$ divides $|H|$. Thus, $|H| = 42n$. We require that $\tfrac{420}{42n} = \tfrac{10}{n}$. This means $n=1,2,5,10$.

4. For number 1 is the ord(a)= gcd(18, 12)=6?

For number 3 how does [(134)(1562)]^34=(134)^34(1564)^34=(134)(1564)^2 show sigma^38. Im very confused about this hint and isn't the right of 6 suppose to be 2?

5. Originally Posted by Luck of the Irish
For number 1 is the ord(a)= gcd(18, 12)=6?
If the order of a were 6, it could not possibly generate a group of order 36! In any case, if ord(a)= 6, then a^12= (a^6)^2= e^2= e which you know is not true. Saying that a^12 and a^18 are not e tells you that the that the order of a cannot be a divisor of either of those: it cannot be 3, 4, 6, 9, 12, or 18 but it MUST divide 36! What does that tell you?

For number 3 how does [(134)(1562)]^34=(134)^34(1564)^34=(134)(1564)^2 show sigma^38. Im very confused about this hint and isn't the right of 6 suppose to be 2?
Perhaps he simply misread "38" as "34". (Or perhaps he wanted to hint at how you should be thinking without doing the problem for you!) The point is that cycling (134) 3 times gives the identity: 3 divides 38 12 times with remainder 2. Since each of those 12 times gives the identity, (134)^38= (134)^2. Similarly cycling (1562) 4 times gives the identity. 4 divides into 38 9 times with remainder 3: (1562)^38= (1562)^3