document.write( "Question 1186054: This for a Abstract Algebra class\r
\n" ); document.write( "\n" ); document.write( "Assume that φ is a homomorphism from the group G to G′. Prove that if K is any subgroup of G′, then φ−1(K) = {a ∈ G | φ(a) ∈ K} is a subgroup of G. \n" ); document.write( "

Algebra.Com's Answer #817043 by robertb(5830) $\"\"$ $\"About$

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Let a, b ∈ $\"phi%5E%28-1%29%28K%29\"$ \r
\n" ); document.write( "\n" ); document.write( "===> $\"phi%28a%29\"$ , $\"phi%28b%29\"$ ∈ K.\r
\n" ); document.write( "\n" ); document.write( "Now $\"phi%28a%29%2Aphi%28b%29+=+phi%28a%2Ab%29\"$ ∈ K because $\"phi\"$ is a homomorphism.\r
\n" ); document.write( "\n" ); document.write( "===> $\"a%2Ab\"$ ∈ $\"phi%5E%28-1%29%28K%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "Now let a ∈ $\"phi%5E%28-1%29%28K%29\"$ . ===> $\"phi%28a%29\"$ ∈ K.\r
\n" ); document.write( "\n" ); document.write( "===>

since $\"phi\"$ is a homomorphism.\r
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\n" ); document.write( "\n" ); document.write( "===> $\"%28phi%28a%29%29%5E%28-1%29+=+phi%28a%5E%28-1%29%29\"$ ∈ K by the uniqueness of the inverse element of G'.\r
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\n" ); document.write( "\n" ); document.write( "===> $\"a%5E%28-1%29\"$ ∈ $\"phi%5E%28-1%29%28K%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Since for any element a, b ∈ $\"phi%5E%28-1%29%28K%29\"$ , a*b and $\"a%5E%28-1%29\"$ ∈ $\"phi%5E%28-1%29%28K%29\"$ , it follows that $\"phi%5E%28-1%29%28K%29\"$ is a subgroup of G.\r
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\n" ); document.write( "\n" ); document.write( "The conclusion can also be obtained by showing that if a, b ∈ $\"phi%5E%28-1%29%28K%29\"$ , then $\"a%2Ab%5E%28-1%29\"$ ∈ $\"phi%5E%28-1%29%28K%29\"$ . The arguments are similar to above.
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