Tools of algebra (3): Groups

Algebra Level 5

How many of the following statements is/are true?

  1. Any cyclic group is an abelian group.

  2. Any simple abelian group is a cyclic group.

  3. Let G G be a finite group, then the following are equivalent:

    a) G |G| is prime.

    b) G G and { e } \{e\} are the only subgroups in G G , and G { e } G \neq \{e\} .

    c) G G is a cyclic group and G Z p G \cong \mathbb{Z}_p for some p p prime.


Relevant notes:

  1. e e denotes the identity element in G G
  2. Abelian and ciclyc groups
  3. Lagrange's Theorem
  4. Normal subgroups
  5. Simple groups
2 3 0 1

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

a) A cyclic group G G is a group generated by a single element. Let a G a \in G such that a = G \langle a \rangle = G , then b , c G , m , n Z \forall \space b,c \in G, \space \exists \space m,n \in \mathbb{Z} such that b = a m b = a^m and c = a n b c = a m a n = a n a m = c b c = a^n \Rightarrow b\cdot c = a^m \cdot a^n = a^n \cdot a^m = c \cdot b \Rightarrow G is an abelian group.

b) Each subgroup in an abelian group G G is a normal subgroup,then if G G is a simple abelian group and G = { e } G = \{e\} , G G is a cyclic group. If G { e } G \neq \{e\} , take a G a \in G with a e a \neq e . Then the subroup generated by a a , thus a \langle a \rangle is a normal subgroup in G G and a { e } \langle a \rangle \neq \{e\} and being G G a simple group, this implies that a = G \langle a \rangle = G , and therefore G G is a cyclic group.

c)

c 1 ) c 2 ) c1) \Rightarrow c2) .- Apply Lagrange's theorem.

c 2 ) c 3 ) c2) \Rightarrow c3) .- If G G is a finite group with { e } \{e\} and G G being the only subroups,we can take a e a \neq e since G { e } G \neq \{e\} and then a \langle a \rangle is a subgroup of G { e } a = G G\neq \{e\}\Rightarrow \langle a \rangle = G \Rightarrow G G is a cyclic group and being G G a finite group G Z p G \cong \mathbb{Z}_p with p p a prime (applying Lagrange's theorem again)

c 3 ) c 1 ) c3) \Rightarrow c1) .- Trivial

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...