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Try constructing a table for a group of order 4 in which the elements are called e,a,b and C and are written in that order along the top row. Show that there are 5 such tables.

I've got 4, idk how they got 5.

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Reply 1
Original post by Inert1a
Try constructing a table for a group of order 4 in which the elements are called e,a,b and C and are written in that order along the top row. Show that there are 5 such tables.


Are you sure that that was asked? AFAIK there are only two groups of order 4 up to isomorphism.
Original post by Zacken
Are you sure that that was asked? AFAIK there are only two groups of order 4 up to isomorphism.
There are distinct possible labellings for the same isomorphic group though. E.g. in Z4 we can have a =1, b=2, c=3 or a=1, b=3, c=2.
Reply 3
Original post by DFranklin
There are distinct possible labellings for the same isomorphic group though. E.g. in Z4 we can have a =1, b=2, c=3 or a=1, b=3, c=2.


Right, I see. Do we actually count those as "distinct tables"? [In that case, I can only come up with 4 as well: labelling elements in the style Z/4Z, Klein-4, and roots of unity]
Original post by Inert1a
Try constructing a table for a group of order 4 in which the elements are called e,a,b and C and are written in that order along the top row. Show that there are 5 such tables.

I've got 4, idk how they got 5.
I also only see 4, and I also did a computer search which came to the same conclusions (doesn't mean I didn't code it wrong, of course).

As Zacken says, there are 2 groups of order 4 up to isomorphism.

If we first look at the cyclic group {e, p, p^2, p^3} (with p^4 = e), there are 6 ways we can assign {a, b, c} to {p, p^2, p^3}. But we'd kind of expect to be able to swap p and p^3 without making a difference and this turns out to be the case.

1abc
abc1
bc1a
c1ab is generated by a=p,b=p^2,c=p^3 and also a=p^3, b=p^2, c=p

1abc
ac1b
b1ac
cba1 is generated by a=p, b=p^3, c=p^2 and by a=p^3, b=p, c=p^2

1abc
a1cb
bca1
cb1a is generated by a=p^2, b=p, c=p^3 and by a=p^2, b=p^3, c=p

These are all the possibilities for this group.

For the Klein 4-group e, p, q, pq, with p^2=q^2=(pq)^2 = e, all possibilities assignments of {a,b,c} to {p,q,pq} give the same labelling:

1abc
a1cb
bc1a
cba1

There's some wiggle room in that the question doesn't specifiy the row order, but if you allow arbitrary row orders there's obviously a lot more than 5 solutions. So I've no idea where they're finding this fifth group either :confused:
(edited 7 years ago)
Original post by Zacken
Right, I see. Do we actually count those as "distinct tables"? [In that case, I can only come up with 4 as well: labelling elements in the style Z/4Z, Klein-4, and roots of unity]
I believe that's the question's intent. It's not exactly something you normally do with groups, however... (Although I believe considerations like this are sometimes important when considering classification of groups. Way beyond the scope of the OP, of course...)
Reply 6
Original post by DFranklin
I believe that's the question's intent. It's not exactly something you normally do with groups, however... (Although I believe considerations like this are sometimes important when considering classification of groups. Way beyond the scope of the OP, of course...)


Okay thank you ever so much for the clarification. I suppose it's a mistake in the question. Would you be able to help me with some other questions that I'm having difficulty with? 
Reply 7
Original post by Inert1a
Okay thank you ever so much for the clarification. I suppose it's a mistake in the question. Would you be able to help me with some other questions that I'm having difficulty with? 


Post them?
Reply 8
Any help would be greatly appreciated.
Reply 9
Original post by Inert1a
Any help would be greatly appreciated.


For Q4) show that the set under addition is a group. This is just basic checking group axioms. Then show that there exists an element a, such that any element in the group is of the form a^k where a^k means a + .. + a = a * k.

For Q5) pretty much the same thing sans having to prove it's a group.
Reply 10
Original post by Zacken
For Q4) show that the set under addition is a group. This is just basic checking group axioms. Then show that there exists an element a, such that any element in the group is of the form a^k where a^k means a + .. + a = a * k.

For Q5) pretty much the same thing sans having to prove it's a group.


Thank you so much! I had a few more from the same text book. They are beyond my specification, but I still wanted to try and complete them.
Reply 11
Original post by Inert1a
Thank you so much! I had a few more from the same text book. They are beyond my specification, but I still wanted to try and complete them.


(as an aide: Q9 is something you know already - look at the way the composition works, now think back to multiplication of complex numbers, notice anything?)

But yeah, this is still just straightforward checking the group axioms, for Q9 again:

(i) is there an identity? - note that (1,0) [this is basically 1 + 0i from the interpretation I mentioned earlier] acts as the identity since (a,b) . (1,0) = (a,b) and the other way around.

(ii) does (a,b) have an inverse? Well, can you find (c,d) such that (a,b) . (c,d) = (1,0) always? This should come down to solving a pair of simultaneous equations ad + bc = 1 and bd - ac = 0 where you treat a and b as constans and c and d as the variables. [this is where we have that (a,b) can't be (0,0) since 0+0i can't have an inverse in complex multiplication]

(iii) Is it associative? Basic boring algebraic bash.

(iv) closure is also straightforward

Q10 is much the same, just check the group axioms - this is actually known as a "direct product", if you want to google and look into it more. Post back if you're having trouble checking the axioms.
Reply 12
Original post by Zacken
(as an aide: Q9 is something you know already - look at the way the composition works, now think back to multiplication of complex numbers, notice anything?)

But yeah, this is still just straightforward checking the group axioms, for Q9 again:

(i) is there an identity? - note that (1,0) [this is basically 1 + 0i from the interpretation I mentioned earlier] acts as the identity since (a,b) . (1,0) = (a,b) and the other way around.

(ii) does (a,b) have an inverse? Well, can you find (c,d) such that (a,b) . (c,d) = (1,0) always? This should come down to solving a pair of simultaneous equations ad + bc = 1 and bd - ac = 0 where you treat a and b as constans and c and d as the variables. [this is where we have that (a,b) can't be (0,0) since 0+0i can't have an inverse in complex multiplication]

(iii) Is it associative? Basic boring algebraic bash.

(iv) closure is also straightforward

Q10 is much the same, just check the group axioms - this is actually known as a "direct product", if you want to google and look into it more. Post back if you're having trouble checking the axioms.


I've understood the relation to complex numbers, and everything in q9 apart from the inverse. I can't seem to find an inverse. Also, would the identity be (0,1) since (b+ai)(d+ci) = (bd-ac) + (ad+bc)i. so the the question basically has it in the format (complex part, real part) ?

If you could also give a quick run through of question 10 that would be greatly appreciated again. Thank you so much. I'm currently on a gap year resitting fp3, don't really have a teacher, so your help REALLY means a lot!
Reply 13
Original post by Inert1a
I've understood the relation to complex numbers, and everything in q9 apart from the inverse. I can't seem to find an inverse. Also, would the identity be (0,1) since (b+ai)(d+ci) = (bd-ac) + (ad+bc)i. so the the question basically has it in the format (complex part, real part) ?


Oops yeah, I just automatically assumed the format was (real part, complex part) - you're correct. The identity is (0,1).

In the same vein, if you have a complex number b + ai, then surely the inverse would be 1/(b+ai) which you can then turn into the form real part + i Im part by multiplying by b - ai / b - ai. This will give you the form of the inverse. Can you see what the inverse is now and prove that it is indeed the inverse?

Spoiler



If you could also give a quick run through of question 10 that would be greatly appreciated again. Thank you so much. I'm currently on a gap year resitting fp3, don't really have a teacher, so your help REALLY means a lot!


Closure: since G is a group and H is a group then g1g2Gg_1 g_2 \in G and h1h2Hh_1 h_2 \in H so (g1g2,h1,h2)G×H(g_1 g_2, h_1, h_2) \in G \times H.
Identity: For any (g,h)G(g,h) \in G we have (eG,eH)×(g,h)=(eGg,eHh)=(g,h)(e_G, e_H) \times (g,h) = (e_G g, e_H h) = (g,h) since eGg=ge_G g = g by group axioms in GG and eHh=he_H h = h by group axioms in HH.
Inverse: Can you check that (g1,h1)(g^{-1}, h^{-1}) is the inverse to (g,h)(g,h)?
Associativity; Can you check that associativity follows from G and H being associative?

In case you're confused; this is saying take two groups G and H, form a new group G * H whose elements are (g,h) for every possible combination g,h. Then the composition in this new group G*H is done by taking the elements of the form (a,b) and (c,d) and composing them by combining the first two elements a and b as you would have done in G and the second two elements c and d as you would have done in H. So (ab, cd). Note that the two operations ab and cd may be different operations, if G and H have different operations.
Reply 14
Original post by Zacken
Oops yeah, I just automatically assumed the format was (real part, complex part) - you're correct. The identity is (0,1).

In the same vein, if you have a complex number b + ai, then surely the inverse would be 1/(b+ai) which you can then turn into the form real part + i Im part by multiplying by b - ai / b - ai. This will give you the form of the inverse. Can you see what the inverse is now and prove that it is indeed the inverse?

Spoiler





Closure: since G is a group and H is a group then g1g2Gg_1 g_2 \in G and h1h2Hh_1 h_2 \in H so (g1g2,h1,h2)G×H(g_1 g_2, h_1, h_2) \in G \times H.
Identity: For any (g,h)G(g,h) \in G we have (eG,eH)×(g,h)=(eGg,eHh)=(g,h)(e_G, e_H) \times (g,h) = (e_G g, e_H h) = (g,h) since eGg=ge_G g = g by group axioms in GG and eHh=he_H h = h by group axioms in HH.
Inverse: Can you check that (g1,h1)(g^{-1}, h^{-1}) is the inverse to (g,h)(g,h)?
Associativity; Can you check that associativity follows from G and H being associative?

In case you're confused; this is saying take two groups G and H, form a new group G * H whose elements are (g,h) for every possible combination g,h. Then the composition in this new group G*H is done by taking the elements of the form (a,b) and (c,d) and composing them by combining the first two elements a and b as you would have done in G and the second two elements c and d as you would have done in H. So (ab, cd). Note that the two operations ab and cd may be different operations, if G and H have different operations.



Okay I've fully understood Q9 now. I understand all the group axioms for question 10. Where you've said, 'combining the first two elements a and b as you would have done in G and the second two elements c and d as you would have done in H. So (ab, cd).' , has confused me slightly. Would it not be (ac, bd) since if you have two elements (g1,h1) and (g2,h2) you end up with (g1g2, h1h2)? Sorry if this seems silly.
Reply 15
Original post by Inert1a
Okay I've fully understood Q9 now. I understand all the group axioms for question 10. Where you've said, 'combining the first two elements a and b as you would have done in G and the second two elements c and d as you would have done in H. So (ab, cd).' , has confused me slightly. Would it not be (ac, bd) since if you have two elements (g1,h1) and (g2,h2) you end up with (g1g2, h1h2)? Sorry if this seems silly.


Urgh, yes - you're fully correct. I wrote it up in too much of a rush
Reply 16
Original post by Zacken
Urgh, yes - you're fully correct. I wrote it up in too much of a rush


Now worries, Your explanations did really help me though, so thanks again! If I do have any more Qs I'll know who to ask :smile:
Reply 17
Original post by Zacken
Urgh, yes - you're fully correct. I wrote it up in too much of a rush


Awesome!
Reply 18
Hi again, I stumbled across the following question on spiral enlargement when looking at complex numbers.

A small snail starts at the origin of an Argand diagram and walks along the real axis for an hour, covering a distance of 8 metres. At the end of each hour it changes its direction by pi/2 anti clockwise; and in each hour it walks half as far as it did in the previous hour.

Find where it is
a) after 2 hours
b) after 4 hours
c) after 8 hours
d) eventually


I can work out the answers individually, but im struggling to find the general form. Is there even a general form? Thanks.
Original post by Inert1a
Hi again, I stumbled across the following question on spiral enlargement when looking at complex numbers.

A small snail starts at the origin of an Argand diagram and walks along the real axis for an hour, covering a distance of 8 metres. At the end of each hour it changes its direction by pi/2 anti clockwise; and in each hour it walks half as far as it did in the previous hour.

Find where it is
a) after 2 hours
b) after 4 hours
c) after 8 hours
d) eventually


I can work out the answers individually, but im struggling to find the general form. Is there even a general form? Thanks.


Shouldn't this be a new thread? Anyway, it looks like you are summing:

S=8(1+eiπ/22+ei2π/222+ei3π/223+)S=8(1+\frac{e^{i\pi/2}}{2}+\frac{e^{i2\pi/2}}{2^2}+\frac{e^{i3\pi/2}}{2^3}+\cdots)

which is a GP.

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