Question

Let G be an additive group and supposed that a e G. Show that for any integers m and n, ma + na = (m+n) a.

[Note: I've already used induction to show that it works for a fixed positive integer n and a nonnegative m. What I'm having trouble with is showing it for negative n. I have, in a previous problem, proven that -(na) = n(-a), and I feel like I"m supposed to use that fact, but I'm not sure how.]

Answer 1

what does a e G mean?

Answer 2

A is an element of G... \[a \in G\]

Answer 3

I thought I had it there, but alas it was not to be.

Answer 4

sigh, u typed so much i thought u were going to post the answer lol, i'll see if i can get one of the gurus or legends to solve this

Answer 5

How familiar are you with rings?

Answer 6

I'm decently familiar with them. My book covered rings and fields before it covered groups.

Answer 7

Ok. So you've proved that
A. for m, n > 0, ma + na = (m + n)a
B1. for m < 0, ma = (-m)(-a)
B2. for m > 0, (-m)a = m(-a)

Yes?

Answer 8

I've proven A (though for m >= 0), and I've proven B2, but not B1... though B1 follows right from B2.

Answer 9

Are you given that G is also an abelian group? Or is that not given?

Answer 10

@KingGeorge: Nope.

Answer 11

The point is, once you have these results, you can extend the range of m and n to negative integers. (And no, G need not be Abelian)

Answer 12

Too bad. If it was abelian I would have had a great answer.

Answer 13

I've been trying different ways of rearranging things such that I can build off m,n > 0, but nothing has stuck, basically.