Abstract Algebra Help

[Avé]

I'm like two steps away from finishing a proof dealing with isomorphisms, but I am unable to explain the following proposition that would allow me to continue. Could someone explain to me how this works:

If gcd(a,b) = 1, a|c, and b|c, then ab|c.

I see that gcd(a,b) = 1 implies there exists x and y such that ax + by = 1. Then multiplying by c gives us cax + cby = c. Since a|ax and b|c, then ab|cax. Since b|by and a|c, then ab|cby. ...

How do I proceed from here? Is it logical to say that because ab|cax and ab|cby, then ab|c? It just doesn't feel right to me.

[Blubbartron]

If a and b are relatively prime, and both divide some number c, then use the definition of divides. a|c, so there exists some k s.t. c = a*k. Similarly for b|c (there exists l s.t. c = b*l). Now divide the latter by a. k = (b*l)/a. Because b and a are relatively prime, then a|l, because k must be an integer by definition. Therefore k = b*j, where j is an integer. Therefore c = a*b*j = (ab)*j and therefore ab|c.

[Woozie]

Since gcd(a,b) = 1, then a does not divide b.

b|c => b*y = c for some y. Since a divides c, then a divides b*y. Since a divides b*y, but a does not divide b, then a divides y.

Since by = c, then b = c/y. So c/ab = cy/ac = y/a, which is an integer because a divides y. Thus, ab|c

[Blubbartron]

I feel all warm and fuzzy for beating Woozie, the resident super-genius in mathematics, to this proof. ^^b Yay math and shit.

[Artemas]

If you two had offspring, it could be Nerd Prime, maybe.

[Avé]

Thanks, I appreciate the help.

[Woozie]

I feel all warm and fuzzy for beating Woozie, the resident super-genius in mathematics, to this proof. ^^b Yay math and shit.

I blame the liberal media and Obama's Health care bill for slowing my response time. A very trustworthy guy on Fox news told me I can blame all my problems on them, and so I will!