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Klein Four Group and Isomorphism Proof

Date: 11/01/2004 at 22:19:13
From: Kate
Subject: Prove Isomorphism

Let |G| = 4.  Prove that either G is isomorphic to C4, or G is 
isomorphic to V.  What is the group V(Klein four group)?

G is given to have order 4, thus it consists of 4 elements.  C4 is the
set of rotations of order 4.  I'm not sure what V is.  Now that we
know the orders match we have to verify that there is a one-to-one
correspondence, but I'm not sure how to do this.  Could you tell me
what V is and how I might go about this proof?  I'm really confused
and I don't know where to start.  Thanks!



Date: 11/04/2004 at 12:22:35
From: Doctor Jordan
Subject: Re: Prove Isomorphism

Hi Kate,

The Klein four group, also called the viergruppe, is the group 

  V={(1), (12)(34), (13)(24), (14)(23)}

Conceptually it is all the group actions on a rectangle such that the
points enclosed by the rectangle are invariant.  The viergruppe is not
cyclic, that is no single element in it generates the whole group.  In
fact, each element is its own inverse, and the composition of any two
non-identity elements is the other non-identity element.

Consider a group G with |G|=4, i.e. G has order 4.  It can either be
cyclic or noncyclic.  In the case that it is noncyclic, of course it
must have an identity element e of order 1.  It must have another
element a, and since G is noncyclic a cannot have order 4, otherwise
<a>=G, that is otherwise G would be cyclically generated by a.  Recall
that the order of an element in a group must divide the order of the
group, so the order of a cannot be 3.  Only the identity element has
order 1.  Thus a has order 2.  Precisely the same argument shows that 
the order 2 elements b and c of G also have order 2.  But a having
order 2 means that aa=e, so a is its own inverse, and the same is true
for b and c. 

But recall that since G is a group it must be closed under its
operation.  Thus, for example, ab is in G.  Since we already have 4
elements, ab must be one of these 4.  It cannot be e, otherwise a 
would be b's inverse and b would be a's inverse, and thus a=b, which 
is false by assumption.  If ab=a or ab=b, then either a or b would 
have to be the identity e, which is also false by assumption.  Thus 
ab=c.  The same argument shows that bc=a. 

Clearly G is Abelian.  Consider the a function f:G->V, defined by
f(e)=(1), f(a)=(12)(34), f(b)=(13)(24), f(c)=(14)(23).  Clearly f is 
an isomorphism between G and V, so G and V are isomorphic.

The other case is where G is cyclic.  Since G is cyclic it must be
Abelian.  Note that some a in G has <a>=G, that is a generates G and a
has order 4.  G must also have the identity e with order 1.  Look at 
the definition of the group C4 at MathWorld here 

  Cyclic Group C4
    http://mathworld.wolfram.com/CyclicGroupC4.html  

and try to finish the proof.  Write me back if you can't get it or 
have any other questions.

- Doctor Jordan, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Modern Algebra

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