Algebra tricky question

Original post by maths10101

Very confused on parts 1 and 2 of this, not sure of the proof at all,
And how do I go abouts part 3?
Thankyou for any help in advance

Can you post a proper picture - this is cut off a little. What are your thoughts on it?

Reply 2

OP

Original post by Zacken

Can you post a proper picture - this is cut off a little. What are your thoughts on it?

Sorry about that and ignore part c, I've figured fhat one out..
For part a, I'm completely clueless of how to carry it out
For part b, would you find the orders of all elements of A^6? And use this to show that g is 1? There must be a quicker way of doing this as this method is quite time consuming

Reply 3

joostan

13

Original post by maths10101

Sorry about that and ignore part c, I've figured fhat one out..
For part a, I'm completely clueless of how to carry it out
For part b, would you find the orders of all elements of A^6? And use this to show that g is 1? There must be a quicker way of doing this as this method is quite time consuming

For a) and b) surely you could just apply Lagrange?
As for b) I wouldn't want to find all the elements of A6, there are 360 of them.

(edited 8 years ago)

Reply 4

OP

Original post by joostan

For a) and b) surely you could just apply Lagrange?
As for b) I wouldn't want to find all the elements of A6, there are 360 of them.

So for a) i've seen that since the cardinality of the subgroup divides the cardinality of the group (S-12), it must be a subgroup by lagrance theorem,,,,does this seem okay?
and for part b....i'm not sure how to prove that

Reply 5

joostan

13

Original post by maths10101

So for a) i've seen that since the cardinality of the subgroup divides the cardinality of the group (S-12), it must be a subgroup by lagrance theorem,,,,does this seem okay?
and for part b....i'm not sure how to prove that

I don't follow what you're saying for a), Lagrange's Theorem states that the order of a subgroup must divide the order of the group.
What is the order of

S_{12}

? What can you then say about the order of the subgroups of

S_{12}

?

As for (b) how many elements of

A_6

are there?
Now if you suppose that an element

g

of order 7 exists, what can you say about the set

1,g,. . .,g^6

?

Reply 6

OP

Original post by joostan

I don't follow what you're saying for a), Lagrange's Theorem states that the order of a subgroup must divide the order of the group.
What is the order of

S_{12}

? What can you then say about the order of the subgroups of

S_{12}

?

As for (b) how many elements of

A_6

are there?
Now if you suppose that an element

g

of order 7 exists, what can you say about the set

1,g,. . .,g^6

?

The order of S_12 is 479001600
and i'm not finding anything about the order of the subgroups online?

Reply 7

Original post by maths10101

The order of S_12 is 479001600
and i'm not finding anything about the order of the subgroups online?

It has order 12! - does 17 divide that? Prove that it doesn't.

Reply 8

OP

Original post by Zacken

It has order 12! - does 17 divide that? Prove that it doesn't.

Ahhh brilliant! Any ideas on part 2 mate?

Reply 9

joostan

13

Original post by maths10101

The order of S_12 is 479001600
and i'm not finding anything about the order of the subgroups online?

You know that the order of

S_{12}

is

12!

.
You want to show that there is no subgroup of order

17

.
If a group of order

n

has a subgroup of order

k

, then we must have

k|n

by Lagrange.
Apply this to this case.

Again, for the second question, the order of an element must divide the order of the group by Lagrange.
Thus if

g^7=1

then what are the possible orders of

g

?

(edited 8 years ago)

Reply 10

For Part 3) use the method of successive squaring.

Reply 11

Original post by fishyfishy1999

For Part 3) use the method of successive squaring.

He's already done part 3. :-)

Reply 12

Original post by Zacken

He's already done part 3. :-)

Oops. The answer is 10 for what it is worth.

Reply 13

Original post by fishyfishy1999

Oops. The answer is 10 for what it is worth.

Given that he's already done it, mind posting an outline of how you do it? I'm interested in knowing and I haven't heard of your method before. If it's not too much trouble, that is.

Reply 14

Original post by Zacken

Given that he's already done it, mind posting an outline of how you do it? I'm interested in knowing and I haven't heard of your method before. If it's not too much trouble, that is.

It's a pain to write down in this box. Just Google Successive Squaring Algorithm for computing a^k mod m.

Quite straightforward.

End up with 18 14 18 14 18 14 18 18 7 10 7 7 10 14 7 10 14 18 10 as the non-zero remainders that are multiplied together to give 439205094557429760000 mod 31 which is 10.

Reply 15

Original post by fishyfishy1999

It's a pain to write down in this box. Just Google Successive Squaring Algorithm for computing a^k mod m.

Quite straightforward.

End up with 18 14 18 14 18 14 18 18 7 10 7 7 10 14 7 10 14 18 10 as the non-zero remainders that are multiplied together to give 439205094557429760000 mod 31 which is 10.

Will do so demain in the morning, thanks.

Reply 16

OP

ImageUploadedByStudent Room1456746198.061610.jpg

Hey, was just wondering if I've done this question correctly for part b?

Posted from TSR Mobile

@zacken

Reply 17

OP

Also, surely Fermat's Little theorem would be easier to apply to work out part c?
@Zacken @fishyfishy1999

Reply 18