Prime power order implies nilpotent: Difference between revisions

Latest revision as of 22:43, 24 June 2009

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group of prime power order) must also satisfy the second group property (i.e., nilpotent group)
View all group property implications | View all group property non-implications
Get more facts about group of prime power order|Get more facts about nilpotent group

Statement

Verbal statement

Any group of prime power order is nilpotent.

Related facts

Similar facts for groups of prime power order

Prime power order implies not centerless: This result is the key ingredient used to prove that any group of prime power order is nilpotent.
Prime power order implies center is normality-large: This is a stronger version of the fact that any group of prime power order is centerless.

Related facts for possibly infinite p-groups

Breakdown for Lie rings

Prime power order not implies nilpotent for Lie rings

Facts used

Prime power order implies not centerless

Proof

We prove the statement by showing that it is possible to construct an upper central series for the group. The proof proceeds by induction on the order of the group. The base case for induction, namely the case of a group of prime order, is clear.

For the induction step, suppose the result is true for all groups whose order is $p^{d},d<r$ . We want to show that the result is true for $p^{r}$ . Let $G$ be a group of order $p^{r},r\geq 1$ . Then, by fact (1), $Z(G)$ is nontrivial, and thus $G/Z(G)$ has order $p^{d}$ with $d<r$ . Thus $G/Z(G)$ is nilpotent, so has an upper central series, and pulling this back gives an upper central series for $G$ .

@@ Line 1: / Line 1: @@
+{{group property implication|
+stronger = group of prime power order|
+weaker = nilpotent group}}
 ==Statement==
@@ Line 16: / Line 20: @@
 * [[p-group not implies nilpotent]]
+===Breakdown for Lie rings===
+* [[Prime power order not implies nilpotent for Lie rings]]
+==Facts used==
+# [[Prime power order implies not centerless]]
 ==Proof==
 We prove the statement by showing that it is possible to construct an [[upper central series]] for the group. The proof proceeds by induction on the order of the group. The base case for induction, namely the case of a group of prime order, is clear.
-For the induction step, suppose the result is true for all groups whose order is <math>p^d, d < r</math>. We want to show that the result is true for <math>p^r</math>. Let <math>G</math> be a group of order <math>p^r, r \ge 1</math>. Then since [[prime power order implies centerless|any nontrivial group of prime power order has nontrivial center]], <math>Z(G)</math> is nontrivial, and thus <math>G/Z(G)</math> has order <math>p^d</math> with <math>d < r</math>. Thus <math>G/Z(G)</math> is nilpotent, so has an upper central series, and pulling this back gives an upper central series for <math>G</math>.
+For the induction step, suppose the result is true for all groups whose order is <math>p^d, d < r</math>. We want to show that the result is true for <math>p^r</math>. Let <math>G</math> be a group of order <math>p^r, r \ge 1</math>. Then, by fact (1), <math>Z(G)</math> is nontrivial, and thus <math>G/Z(G)</math> has order <math>p^d</math> with <math>d < r</math>. Thus <math>G/Z(G)</math> is nilpotent, so has an upper central series, and pulling this back gives an upper central series for <math>G</math>.