{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework7

# Homework7 - Homework 7 Solutions 1 Let |G| = pn for n 0...

This preview shows pages 1–2. Sign up to view the full content.

Homework 7 Solutions 1. Let | G | = p n for n 0. ( ) Suppose G is simple. Since G is simple, n 6 = 0. In addition, we know that G has normal subgroups of order p k for k n . Since G is simple, n must equal 1. Thus | G | = p . ( ) Suppose | G | = p . By Lagrange’s theorem, since p is prime, the only subgroups of G are { e } and G itself. Thus G has no non-trivial, proper normal subgroups. 2. (a) C G ( x ) H = H ∩ { g G | gxg - 1 = x } = { g H | gxg - 1 = x } = C H ( x ) (b) [ G : H ] = 2 H C G . By the second isomorphism theorem we know C G / ( H C G ) = HC G /H and HC G G . In addition, since [ G : H ] = 2 then HC G = H or G . By part (a) we have C G / ( H C G ) = C G /C H = HC G /H . Since | x H | = | H | / | C H | and | C G /C H | = | HC G /H | we have | x H | = | HC G | / | C G | . Since HC G = H or G and | G | = 2 | H | we have | x H | = 1 2 | G | / | C G | or | G | / | C G | and thus | x H | = 1 2 | x G | or | x G | . 3. (a) Assume there exists H G such that [ G : H ] = n > 1 and G is simple. The generalized Cayley’s theorem implies there exists a homomorphism ϕ : G S n with kerϕ H .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

Homework7 - Homework 7 Solutions 1 Let |G| = pn for n 0...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online