Thread: Show a group is isomorphic. Table included!

Let T be the group with four elements e,x,y,z with multiplication table

lllll | e | x | y | z |
l e | e | x | y | z |
l x | x | e | z | y |
l y | y | z | x | e |
l z | z | y | e | x |

Show that this group is isomorphic to Z4.
(Z being all integers)
(e being the identity)

I know I have to define a bijection between these groups which is a homomorphism but then I'm lost.

Z4 is cyclic of order 4. in Z4, 1 is an element of order 4 which generates the group (3 is also a generator).

if T and Z4 are to be isomorphic, T has to have an element of order 4.

well, it's not e, that just leaves 3 to check. what is the order of x?

x^2 = e, so x is of order 2. let's keep looking.....

y^2 = x. this looks promising....

y^3 = (y^2)y = xy = z. so y is not of order 1,2 or 3...what's left? just to be sure, let's check:

y^4 = (y^3)y = zy = e, so yep, y is of order 4.

a homomorphism has to map identity to identity:

so e--->0 is already chosen for us.

an isomorphism has to send elements to images of the same order, so we have two choices for y:

y-->1, or y-->3. let's pick y--->1. we can now write T as {e,y,y^2,y^3} (since we know y^2 = x, and y^3 = z)

now, convince yourself that sending y^k <---> k (mod 4) is an isomorphism.

the bijection part is easy...why is it a homomorphism (can you think of a way to prove it "in general" without verifying all 16 possible products in T)?

Originally Posted by Deveno

Z4 is cyclic of order 4. in Z4, 1 is an element of order 4 which generates the group (3 is also a generator).

if T and Z4 are to be isomorphic, T has to have an element of order 4.

well, it's not e, that just leaves 3 to check. what is the order of x?

x^2 = e, so x is of order 2. let's keep looking.....

y^2 = x. this looks promising....

y^3 = (y^2)y = xy = z. so y is not of order 1,2 or 3...what's left? just to be sure, let's check:

y^4 = (y^3)y = zy = e, so yep, y is of order 4.

a homomorphism has to map identity to identity:

so e--->0 is already chosen for us.

an isomorphism has to send elements to images of the same order, so we have two choices for y:

y-->1, or y-->3. let's pick y--->1. we can now write T as {e,y,y^2,y^3} (since we know y^2 = x, and y^3 = z)

now, convince yourself that sending y^k <---> k (mod 4) is an isomorphism.

the bijection part is easy...why is it a homomorphism (can you think of a way to prove it "in general" without verifying all 16 possible products in T)?

Can this be proven by figuring the table out?
T' =

lllll | 0 | 1 | 2 | 3 |
l 0 | 0 | 1 | 2 | 3 |
l 1 | 1 | 2 | 3 | 0 |
l 2 | 2 | 3 | 0 | 1 |
l 3 | 3 | 0 | 1 | 2 |

A homomorphism between T and T' is any function, f, satisfying f(a*b) = f(a)*f(b) for all a,b in T

For this table (T'), f(2*3) = f(1) vs f(2) * f(3) = f(1), right?

which is the same for T which would be:
f(y*z) = f(x) vs f(y) * f(z) = f(x), right?

yes, you CAN compare tables....but that approach will prove useless with any group of large size (even a group of order 12 takes more time to construct a table than most people are willing to do).

but behold! here is my "magic proof"

define φ(y^k) = k (mod 4). we already have seen this is a bijection.

φ((y^k)(y^m)) = φ(y^(k+m)) = k+m (mod 4) = k (mod 4) + m (mod 4) = φ(y^k) + φ(y^m), no muss, no fuss, φ is a homomorphism.

Originally Posted by Deveno

yes, you CAN compare tables....but that approach will prove useless with any group of large size (even a group of order 12 takes more time to construct a table than most people are willing to do).

but behold! here is my "magic proof"

define φ(y^k) = k (mod 4). we already have seen this is a bijection.

φ((y^k)(y^m)) = φ(y^(k+m)) = k+m (mod 4) = k (mod 4) + m (mod 4) = φ(y^k) + φ(y^m), no muss, no fuss, φ is a homomorphism.

Wow! That proof is delightful. Thank you! Certainly too cool to be in my textbook.