Groups of Prime Power Order

  • #1
Euge
Gold Member
MHB
POTW Director
2,058
215
If ##p## is a prime, and ##G## is a finite non-cyclic ##p##-group, show that there is a normal subgroup ##N## of ##G## such that ##G/N## is isomorphic to ##\mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}##.
 
  • Like
Likes jbergman and topsquark
Physics news on Phys.org
  • #2
Proof by induction on ##k##, where ##|G| = p^k##. The assertion is vacuous when ##k = 1## since every group of prime order is cyclic. Now assume ##k > 1## and the assertion holds for all ##p##-groups of order smaller than ##|G|##.

If ##G## is abelian, then by the fundamental theorem of finitely generated abelian groups, there is an isomorphism ##h## from ##G## onto a direct sum of cyclic ##p##-groups ##\mathbb{Z}/p^{a_1}\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/p^{a_m}\mathbb{Z}##. Note ##m > 1## since ##G## is non-cyclic. If ##N## is the preimage of $$p\mathbb{Z}/p^{a_1}\mathbb{Z} \oplus p\mathbb{Z}/p^{a_2}\mathbb{Z} \oplus \bigoplus_{2 < j \le m} \mathbb{Z}/p^{a_j} \mathbb{Z}$$ under ##h##, then ##N## is a normal subgroup of ##G## such that ##G/N \simeq \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}##.

In case ##G## is nonabelian, ##G/Z(G)## is noncylic. Since the center of a ##p##-group is nontrivial, ##G/Z(G)## has order smaller than ##|G|##. By the induction hypothesis, there is a normal subgroup ##\overline{N}## of ##G/Z(G)## such that ##(G/Z(G))/N \simeq \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}##. By the lattice isomorphism theorem, ##\overline{N} = N/Z(G)## for some normal subgroup ##N## of ##G## containing ##Z(G)##. Finally, the third isomorphism theorem yields ##G/N \simeq (G/Z(G))/\overline{N} \simeq \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}##.
 
  • Like
Likes topsquark

Similar threads

  • Math POTW for University Students
Replies
1
Views
2K
  • Math POTW for University Students
Replies
1
Views
617
Replies
5
Views
962
  • Linear and Abstract Algebra
Replies
1
Views
744
  • Linear and Abstract Algebra
Replies
1
Views
898
  • Linear and Abstract Algebra
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
973
  • Math Proof Training and Practice
3
Replies
80
Views
5K
Back
Top