GATE Overflow Test Series | Mock GATE | Test 5 | Question: 21

Question

Which of the following is/are correct? (Mark all the appropriate choices)

• A. If $G$ is an abelian group of order $36,G$ has an element of order $3$.
• B. If $G$ is an abelian group of order $308,G$ has an element of order $11$.
• C. If $G$ is an abelian group of order $36,G$ has an element of order $5$.
• D. If $G$ is an abelian group of order $308,G$ has an element of order $6$.

Answer 1

In the case of finite abelian groups, the order of any element divides the order of the group and there must be an element in the group corresponding to every prime divisor of the order of the abelian group

Divisors of $36 = 1, 2, 3, 4, 6, 9, 12, 18, 36 = 2^{2} \times 3^{3}.$

Divisors of $308 = 1, 2, 4, 7, 11, 14, 22, 28, 44, 77, 154, 308 = 2^{2} \times 7 \times 11.$

So, the correct answer is $A;B.$

Reference: https://math.stackexchange.com/questions/3572431/let-g-be-a-finite-abelian-group-and-let-p-be-a-prime-that-divides-the-order-of-g

More reference: https://math.stackexchange.com/questions/477500/showing-that-a-finite-abelian-group-has-a-subgroup-of-order-m-for-each-divisor

Comments

yes, it is necessary. https://math.stackexchange.com/questions/3572431/let-g-be-a-finite-abelian-group-and-let-p-be-a-prime-that-divides-the-order-of-g

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@Arjun sir in the above link the proof is about the prime devisor,  whereas the statement in the solution which says the following talks about any devisor prime or non prime, so I think above statement needs an edit. Can you please check.

In the case of finite abelian groups, the order of any element divides the order of the group and there must be an element in the group corresponding to every divisor of the order of the abelian group

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yes, that’s correct. Fixed now.