Hey guys,

I need to prove that there exists 2 groups with order 4 which are not isomorphic.

I am short on examples. I am just surfacing this topic. Can you please help me out?

Printable View

- August 17th 2012, 02:18 PMmathrldIsomorphism
Hey guys,

I need to prove that there exists 2 groups with order 4 which are not isomorphic.

I am short on examples. I am just surfacing this topic. Can you please help me out? - August 17th 2012, 02:53 PMPlatoRe: Isomorphism
Look at the Kline Four Group.

- August 17th 2012, 03:37 PMDevenoRe: Isomorphism
start from first principles.

we have G = {e,a,b,c}.

again, the products involving e are boring. ignore them. we need to focus on a,b and c.

let's start with a. a*a has to be some member of the group.

a*a cannot be a (we've been over this before). that leaves us with 3 choices:

a*a = e

a*a = b

a*a = c

let's look at the first choice: a*a = e (this means a = a^{-1}). so far, so good.

if we adopt this (for the time being), then we can look at a*b.

a*b = e is not possible (for that implies a*a = a*b and thus a = b).

a*b = a is not possible (for that implies b = e).

a*b = b is not possible (for that implies a = e).

so...if a*a = e, it must be the case that a*b = c. hold that thought!

now, i claim that we can always find one of a,b or c such that either:

a*a = e

b*b = e, or

c*c = e.

why? look at inverses:

we know e^{-1}= e...boring!

well, if a*a = e, then we're done, a is it's own inverse. so suppose not. then a^{-1}is either b or c. if b is a's inverse, the a is b's inverse. in that case, c has to be its own inverse (it can't be e,a or b, whose inverses are already spoken for). on the other hand, if c is a's inverse, then a is c's inverse, so b has to be it's own inverse. so no matter how we slice it, at least one of a,b, or c has to be its own inverse.

now, for us, a,b and c are just symbols representing the elements of G. let's re-label them, if we need to, so that a is an element that is its own inverse.

this means we can write G = {e,a,b,a*b}. let's write out all 16 products, filling in the ones we can figure out:

e*e = e

e*a = a

e*b = b

e*(a*b) = a*b

a*e = a

a*a = e <---by the above discussion

a*b = a*b (by definition)

a*(a*b) = (a*a)*b = e*b = b

b*e = b

b*a = ...?

b*b = ....?

b*(a*b) = ...?

(a*b)*e = a*b

(a*b)*a = ...?

(a*b)*b = ...?

(a*b)*(a*b) =...?

as you can see, we already know 10 of the 16. let's work on the remaining 6.

b*a = e implies b = a^{-1}. but a = a^{-1}, and a and b are different. so this can't happen.

b*a = a implies b = e. bad choice.

b*a = b implies a = e. another bad choice.

so b*a has to be a*b. this means (a*b)*a = a*(b*a) = a*(a*b) = (a*a)*b = e*b = b. 2 down, 4 to go.

when we try to evaluate b*b, we hit a dead-end. that's too bad, because if we knew what b*b was, we could also figure out (a*b)*b = a*(b*b), as well as:

b*(a*b) = (b*a)*b = (a*b)*b = a*(b*b).

that would also take care of:

(a*b)*(a*b) = (b*a)*(a*b) = b*(a*a)*b = b*e*b = b*b.

so it all boils down to what b*b is.

b*b = b isn't possible (that would mean b = e, been there, done that. it doesn't work).

b*b = a*b also isn't possible (that would mean b = a, not gonna happen).

so we are left to decide between: b*b = e, and b*b = a. you can play with this a bit yourself, but i'll just "skip ahead": either one works.

suppose b*b = a. this is the same as saying: b^{2}= a. then a*b = b^{2}*b = b^{3}, and we get:

G = {e,b,b^{2},b^{3}}.

note that this is the same as saying: 4 is the smallest positive number n, for which b^{n}= e.

on the other hand, suppose b*b = e. then (a*b)*(a*b) = b*b = e, too...every element of G is its own inverse! in this case, we can't "simplify" G any further:

G = {e,a,b,a*b} is the best we can do.

********

ok, perhaps you see that perhaps two different groups of order 4 might be possible...but does "one of each" actually occur?

well, yes. a group of the "first kind" is:

{0,1,2,3} under addition modulo 4 (in this case, "1" plays the role of "b"). in this group a*b is written a+b, and rather than b^{n}, we write nb. this is just notation, but it often throws people off:

nb does NOT mean "n times b", it means "b+b+...+b" (n times) the *same way*

b^{n}means "b*b*...*b" (n times). the group {0,1,2,3} is called "the cyclic group of order 4" (Z_{4}, when it is written additively, and C_{4}, when multiplicative notation is used).

a group of the second kind is:

{(0,0),(1,0),(0,1),(1,1)} where (a,b) + (a',b') = (a+a' (mod 2), b+b' (mod 2)). this group is called "the klein 4-group", or V, or Z_{2}xZ_{2}, when it is written additively. - August 17th 2012, 03:50 PMmathrldRe: Isomorphism
Wow Thanks. I really appreciate it. You respond it great detail that its easy to follow. I am taking online math courses so I kinda learn from diff problems and examples. I really appreciate it