Group Theory

Suppose G has precisely one subgroup of order 5, and one subgroup of order 7. What is G?

Expert Solution

Again, G = Z35.

Indeed, suppose the subgroup of order 5 is H, and the one of order 7 is K. Then

H ∪ K has

1 + (5 − 1) + (7 − 1) = 11

elements. Choose an element a from G \ (H ∪ K). The order of a is 1, 5, 7 or 35.

We claim the order of a is 35. The order of a is not 1 because it is not the identity.

The order of a is neither 5 nor 7, for otherwise a would generate a subgroup hai of

order 5 or 7, distinct from H or K, and we know there is precisely one subgroup

of order 5 (namely, H), and precisely one subgroup of order 7 (namely, K).

Thus G = <a>, and so G is cyclic.

Nipun Maity answered 2 years ago

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