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I don't know whether I am right or wrong. Can anyone help me to clear below problem ?

Question 1: Let C2 be the cyclic group of order 2 and C202 be the cyclic group of order 202. Find all subgroups of C2 X C202 and the product being defined component-wise.

Answer 1: By the Fundamental Theorem of Cyclic groups, given any cyclic group of order n, there is exactly one cyclic subgroup of order m, for every m such that m|n, and no others. So, finding all of the subgroups of a cyclic group is simply a matter of determining all the factors of the order of that group and coming up with a single subgroup for each one.

C2 has order 2, and the only factors of 2 are 1 and 2. Thus, the only subgroups of C2 are the trivial subgroup and C2itself. C202 has order 202 and the factors of 202 are 1, 2, 101 and 202. Thus, there is the trivial subgroup, C202 itself, and exactly one subgroup of order 2.

Question 2: Suppose I have C15 and C3 X C5 cyclic group. How to prove or disprove that it is isomorphic?

Answer:

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  • $\begingroup$ To me, the answer to the first question is alright. The second one takes a bit of thought, but obvious if you know some lemmas/theorems. $\endgroup$ – NasuSama Feb 17 '14 at 18:17
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First question. You missed the subgroup of order $101$ which is a divisor of $202$.

Secondly, note that $3$ and $5$ are relatively prime. See if you can convince yourself, even better, prove the following VERY useful theorem:

$$ C_{\large mn} \cong C_{\large m}\times C_{\large n}\;\text{ if and only if } \;\gcd(m,n) = 1$$

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