Question

Answer #1

Let G be a finite group of order n, where n is prime.

If possible, let H be a subgroup of order m,say.

Then .

But by Lagrange's theorem m is a divisor of n.

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**Lagrange's Theorem:**

The order of a subgroup of a finite group divides the order of the group.

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Also since n is prime, either m=1 (or) m=n.

(or) H=G.

But these two are improper subgroups of G.

Any group of prime order does not have any proper subgroup.

Thus the total number of subgroups of a group of prime order is 2.

They are and G itself.

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