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1. (a) If you recall the argument we made when finding that all simple finite abelian groups are cyclic of prime order assume
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We first show that G is a finite group.

Suppose that G is a simple abelian group.

Let g∈G be a non identity element of G. Then the group 〈g〉 generated by g is a subgroup of G. Since G is an abelian group, every subgroup is a normal subgroup.

Since G is simple, we must have 〈g〉=G. If the order of g is not finite, then 〈g^2〉 is a proper normal subgroup of 〈g〉=G, which is impossible since G is simple.
Thus the order of g is finite, and hence G=〈g〉 is a finite group.

Let pp be the order of gg (hence the order of G).
Seeking a contradiction, assume that p=m n is a composite number with integers m>1,n>1. Then 〈gm〉 is a proper normal subgroup of G. This is a contradiction since G is simple.

Thus pp must be a prime number.
Therefore, the order of G is a prime number.

Since G is abelian, every subgroup is normal. Since G is simple, the only subgroups of G are 11 and G, and |G|>1, so for some x∈G we have x≠1 and (x〉≤G, so 〈x〉=G. Suppose x has infinite order. Then 〈x^2〉≤G but 〈x^2〉≠〈x〉〈x^2〉≠〈x〉, a contradiction. So x, and therefore G, has finite order. Suppose x has composite order n so for some p>1 that divides n, 〈x^p〉 is a proper non-trivial subgroup of G, so G is not simple. So G is a cyclic group of prime order.

hence we show that G is a simple abelian group and then G is of prime order.

without the finite-ness assumption cant make an argument of consideration. if we estimate sometimes it might give same conclusion.(b) here I reed to paove trat -theas no ComposHion sertes that is, no normal series et SubToufg at 쿤 some neN Since 쿤 ts abel

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