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So I am in week 3 of my intro to group theory class, and I'm lost! My teacher did not go over generators of groups in class. The book is not much help to me either. Any help getting this started would be great!

Show that $\mathbb Z$ is generated by 5 and 7

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Show that every $n \in \mathbb{Z}$ can be written as $n=5m+7k$ with $m,k \in \mathbb{Z}$. – njguliyev Oct 15 '13 at 21:09
Do you know what is meant by "$5$ and $7$ generates $\mathbb{Z}$"? Can you given an English translation of what you think that means? – EuYu Oct 15 '13 at 21:10
From my readings and research the generator is a subset which every element of the group can be expressed as the combination(under the group operation) of its subset and it's inverses. Therefore, if 5 and 7 are generators of $\mathbb z, then $\mathbb Z is the any combination of 5 and 7 and their inverses. The comment above was very helpful in putting that together as well. Is it possible for $\mathbb Z to be generated by any 2 integers? – user101097 Oct 15 '13 at 21:29
@user101097: Is $\mathbb Z$ generated by 2 and 4? – azimut Oct 15 '13 at 21:39
Well, my reasoning here is torn. I want to say no because the greatest common divisor is not 1, it is 2. But, for any combination of 2 and 4 and their inverses will result in an integer, which is in Z. If n is an element of Z, and n=2m+4k with m,k being elements of Z, then all the possible outcomes for n will be in Z. – user101097 Oct 15 '13 at 22:03

Hint: $$5 + 5 + 5 - 7 - 7 = 1$$

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If you are ever given a set of integers $n_1,\ldots,n_k$ whose greatest common divisor is $1$, then you can find an integer combination of the integers which results in $1$. This should be sufficient for your problem and generalizations of it.

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