FP3 Groups

Quick question please!

R: o where a o b = |a-b|

I thought this has an identity element of 0 where a o 0 = 0 o a = a
However, my book says that there is no identity.

Can someone please explain?

(edited 10 years ago)

Reply 1

tallguitarist

Original post by violinmusicfan

Quick question please!

R: o where a o b = |a-b|

I thought this has an identity element of 0 where a o 0 = 0 o a = a
However, my book says that there is no identity.

Can someone please explain?

if a is negative and you blob it with 0 you will not get a. a o 0 = |a|, not a if a is negative. So basically if a is negative there is no element b such that a o b = a

(edited 10 years ago)

OP

Understand, thanks!

OP

Can someone help with this question?
a^4 = e
b^2 = e
a o b = b o a^3

Show that b o (a^2 o b) = a^2.

Reply 4

ghostwalker

17

Original post by violinmusicfan

Can someone help with this question?
a^4 = e
b^2 = e
a o b = b o a^3

Show that b o (a^2 o b) = a^2.

Write it as b o a o a o b, since the group operation is associative, and now find a way to use the information you've been given. You've only got one rule you can use at each stage, so it should be straight forward.

Reply 5

violinmusicfan

OP

1

Thanks!
When do you know when you can write a^2 (a x a) as a o a.
o (blob) can refer to any binary operation right? (can also be addition)

Reply 6

Mark13

Original post by violinmusicfan

Thanks!
When do you know when you can write a^2 (a x a) as a o a.
o (blob) can refer to any binary operation right? (can also be addition)

a^2 is just shorthand for a o a, so you can write it whenever you want, as long as it's clear what the notation means.

In practice, the binary operation of a group is usually written in the same way we write multiplication (so a o b is just written as ab), or in the same way we write addition (so a o b is written a+b). So, just as in actual multiplication, we use a^2 as short hand for aa, we can use a^2 as shorthand for aa in the context of groups as well.

Reply 7

violinmusicfan

OP

1

Thanks for the inputs! Very clear

Reply 8

violinmusicfan

OP

1

Besides finding orders of the elements of both groups,

how could one show that (Z13 - {0}, x mod 13 is isomorphic to (Z12, +)?

Reply 9

ghostwalker

17

Original post by violinmusicfan

Besides finding orders of the elements of both groups,

how could one show that (Z13 - {0}, x mod 13 is isomorphic to (Z12, +)?

Finding the order of each element will not show that the two groups are isomorphic.

You need to construct an isomorphism between the two.

Reply 10

violinmusicfan

OP

1

Sorry but how do you construct an isomorphism exactly?
The book says the function f: G -> H is called an isomorphism which I don't really understand and know how to apply.
Should I begin by drawing out the Cayley table?

Reply 11

Farhan.Hanif93

18

Original post by violinmusicfan

Sorry but how do you construct an isomorphism exactly?
The book says the function f: G -> H is called an isomorphism which I don't really understand and know how to apply.
Should I begin by drawing out the Cayley table?

Assuming this is OCR FP3, a Cayley table argument is probably what is expected.

OP

Ok, thanks!

OP

Are subgroups always isomorphic to their groups?

Reply 14

Noble.

17

Original post by violinmusicfan

Are subgroups always isomorphic to their groups?

No.

Reply 15

Mark13

Original post by violinmusicfan

Are subgroups always isomorphic to their groups?

No - remember that two groups G and H are isomorphic if there exists a bijective homomorphism (called an isomorphism) between them. If H is properly contained in G, then there no bijective maps from H to G, let alone homomorphisms.