1. If p is an odd prime then there are no groups with exactly p elements of orde

Question

1. If p is an odd prime then there are no groups with exactly p elements of order p.

2. Let a and be group elements such that |b|=2 and bab=a^4. The possible orders of a are 1,3, and 5.

3. If G is a group with exectly 8 elements of order 10, then it has exactly 8 cyclic subgroups of order 10.

4. If R is a ring such that for every b element R, b=b^2, then R is commutative.

5. If U= {(x,y,z element R, x= 2y+3z}, then U is a subspace of R^3.

6. Let H = {g^2 : g element G} for some group G. H subgroup of G

Explanation / Answer

1)

So say p is an odd prime and G is a group with exactly p elements of order p. Let g be one of those elements. Then we know that the cyclic subgroup generated by g has p - 1 generators, namely g, g2, ... , gp - 1, each of which has order p. Thus there is one more element of G of order p, say x, and it cannot be a power of g. But this means x, x2, ... , xp - 1 all have order p, and all generate the same subgroup. Thus the sets {g, g2, ... , gp - 1} and {x, x2, ... , xp - 1} are disjoint, so G must have at least 2p - 2 elements of order p. Since p is an odd prime, hence bigger than 2, 2p - 2 is bigger than p, which is a contradiction. (What this argument really shows is that the number of elements of G of order p is a multiple of p - 1. Since no odd prime is a multiple of p - 1, G can't have exactly p elements of order p.)

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