Klein Four Group and Isomorphism ProofDate: 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/ |
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