Well since g=g'.Multiply both sides by (g')^-1 and you'll get the right hand side equal e'9the identity in G' and the left is g.(g')^-1, an element that's in G mapping to the identity so it's in the kernel.

Then he says 'so that g.(g')^-1 is in the kernel.' Why you might ask, as I did. In his review of the second exam, one of the questions deal with a homomorphism phi from a group where|G|>N! to a group of order N! must have a kernel.He goes on to say there must be mappings of distinct g and g' in G such that map to the same element in G'. He also reviews his exams which he acknowledges were too difficult. For more information about the MCAT exam, visit Gross's lectures are very complete. The content is also included in the Pre-health Collection within MedEdPORTAL’s iCollaborative sponsored by the AAMC: *MCAT® is a program of the AAMC and related trademarks owned by the Association include Medical College Admission Test, MCAT, and MCAT2015. These videos, questions, and articles are available under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License (available at ). All materials are categorized according to the pre-health competencies tested by the new MCAT exam however, the content in this collection is not intended to prescribe a program of study for the new MCAT exam. All content in this collection has been created under the direction of the Khan Academy and has been reviewed under the direction of the AAMC (Association of American Medical Colleges).

Content will be added to the collection through 2015. The collection contains more than 1000 videos and 2800 practice questions. MCAT This collection is being developed for the revised MCAT® exam that will first be administered in April 2015.

Algebra 1 khan