abelian p-group not divisible - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-24T01:04:24Z http://mathoverflow.net/feeds/question/67888 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67888/abelian-p-group-not-divisible abelian p-group not divisible stacy 2011-06-15T19:25:16Z 2011-06-15T19:56:20Z <p>why if G is an abelian p-group not divisible then exists an element g in G which is not divisible by p? thanks</p> http://mathoverflow.net/questions/67888/abelian-p-group-not-divisible/67890#67890 Answer by Richard Rast for abelian p-group not divisible Richard Rast 2011-06-15T19:56:20Z 2011-06-15T19:56:20Z <p>As Pace said, but with more detail:</p> <p>If \$G\$ is an abelian \$p\$-group, then for any \$g\$ in \$G\$, the order of \$g\$ is a power of \$p\$, say \$p^k\$. Thus for any integer \$n\$ coprime with \$p\$, \$n\$ is a unit (mod \$p^k\$), so for some \$m\$, \$nm=1\$ mod \$p^k\$. So \$n(mg)=(nm)g=(ap^k+1)g=g+a(p^kg)=g+0=g\$. Thus \$g\$ is divisible by \$n\$.</p> <p>This holds for any \$g\$; so if every \$g\$ is divisible by \$p\$, they are also divisible by \$p^n\$ for all \$n\$, so they are all divisible by \$p^nk\$ for any \$n\$ and any \$k\$ coprime to \$n\$, which is to say, any nonzero number.</p>