What is the normal subgroup of order 125 in a group of order (35)^3?

[ehrenfest]

[SOLVED] group of order (35)^3

Homework Statement

Show that every group of order (35)^3 has a normal subgroup of order 125.

Homework Equations

The Attempt at a Solution

I was trying to use the Third Sylow Theorem to show that there can only be one Sylow 5-subgroup but 21 Sylow 5-subgroups seems to work fine.

 

[jacobrhcp]

35=5x7
35^3=(5^3=125) x 7^3

now use fundamental theorem of arithmetic and first and second sylow theorem, and the third too.

 

[ehrenfest]

How could the fundamental theorem of arithmetic possibly be useful?
By the first sylow theorem, there exists a Sylow p-group of order 125. I need to show that there can only be one such subgroup. The third Sylow theorem says that the number of Sylow 5-groups divides 35^3 and is congruent to 1 mod 5. 21 divides 35^3 and is congruent to 1 mod 5. So what am I doing wrong?

 

[quantumdude]

21 does not divide 35^3. How could it? 35 has no factor of 3. So there, the Fundamental Theorem of Arithmetic is useful here after all!

 

[ehrenfest]

Sorry. We need to show that 5n+1 divides 35^2 only if n=0. Assume n is greater than 0. 35^2=(5^2)*(7^2) so when is 5n+1 a factor of (5^2)*(7^2)? Only when it is a factor of 7^2? And it is never a factor of 7^2. Therefore there is only one Sylow 5-group which must then be normal by the Second Sylow Theorem. Is that right?

 

[quantumdude]

Your last post confuses me. I think you got it, so I'm just going to post the complete solution.

Please don't do that. When you registered you were presented with the PF Global Guidelines right before you clicked the button that says, "I agree."

Here is a relevant excerpt.

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.[/color]

Color added for emphasis.

 

[jacobrhcp]

sorry, I knew and read that, but I saw no harm as he already showed his work and I thought he got it, but just needed to get it straight. You're right to delete it though.

ehrenfest: what you wrote is about right. Can you see why 1 mod 5 never divides 7^3?

 

[ehrenfest]

Yes. I actually saw the deleted post in the e-mail notification and it was exactly the same proof I had, except perhaps more clear. :)