# Finite abelian p-group

• Dec 3rd 2011, 11:37 AM
dwsmith
Finite abelian p-group
Suppose G is a finite abelian p-group. Prove that if G is cyclic iff G has exactly p-1 elements of order p.

$\forall g\in G, \ \ g^{p^{\alpha}}=1, \ \ |G|=p^{\alpha}$

$(\Rightarrow)$
Suppose G is cyclic. Then for a $g\in G$, $G=$.

What next? I am at a loss.
• Dec 3rd 2011, 02:49 PM
Drexel28
Re: Finite abelian p-group
Quote:

Originally Posted by dwsmith
Suppose G is a finite abelian p-group. Prove that if G is cyclic iff G has exactly p-1 elements of order p.

$\forall g\in G, \ \ g^{p^{\alpha}}=1, \ \ |G|=p^{\alpha}$

$(\Rightarrow)$
Suppose G is cyclic. Then for a $g\in G$, $G=$.

What next? I am at a loss.

If $G$ is cyclic, you know it's isomorphic to $\mathbb{Z}_{p^\alpha}$ how many elements o order $p$ does this have (hint:recall cyclic groups have only one subgroup of a given order)? Assume now that $G$ has only $p-1$ elements of order $p$. Choose $g\in G$ with $|g|=p$, now assume that $x\in G-\langle g\rangle$, you know that $\langle x\rangle$ intersects trivially with $\langle g\rangle$ (why?) and so $\langle x\rangle$ is a cyclic subgroup of $G$ whose order is divisible by $p$--so?
• Dec 3rd 2011, 02:53 PM
Deveno
Re: Finite abelian p-group
if G is cyclic, it has only one subgroup of order p. since any element of order p generates a subgroup of order p, all elements of order p must generate the same subgroup of G (since G has only ONE subgroup of order p, because its cyclic). so all elements of order p must all lie in that one subgroup of order p, which has (of course) p-1 elements of order p.

now suppose that G has exactly p-1 elements of order p. pick one of them, say x, we have a cyclic subgroup of G of order p <x>. this is the ONLY subgroup of order p of G.

now suppose H is a subgroup of order p^2, which has to be cyclic, because the direct product of 2 subgroups of order p is not possible. let y be a generator of H. then y^p has order p, so y^p is in <x>. hence <x> is a subgroup of H.

now suppose that K is a subgroup of order p^3....can you see where this is going? show that any subgroup of order p^k has to be cyclic, or else it contains more than one subgroup of order p.

conclude that G must be cyclic.