By Thomas C. Craven

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**Extra info for Abstract algebra [Lecture notes]**

**Example text**

This illustrates a major theorem that we have just proved. 26 (Lagrange’s Theorem). If K is a subgroup of a finite group G, then the order of K divides the order of G. Indeed, |G| = |K|[G : K]. 12 The proof was only a matter of noticing that the cosets are the equivalence classes of an equivalence relation and they all have the same size! At least as useful as this theorem is its corollary obtained by applying it to the cyclic subgroups generated by elements of G. 27. Let G be a finite group. Then (1) The order of every element divides the order of the group.

10. Let f : R → S be any homomorphism of rings and let K = ker f . Then K is an ideal in R. Proof. We know 0 ∈ K, so K = ∅. Let a, b ∈ K. Then f (a) = f (b) = 0, so f (a − b) = f (a) − f (b) = 0. For any r ∈ R, f (ra) = f (r)f (a) = f (r) · 0 = 0. Similarly, f (ar) = f (a)f (r) = 0. Thus a − b, ra and ar are also in K, hence K is an ideal. And furthermore, the kernel tests for injectivity just as it does for linear transformations. 11. Let f : R → S be any homomorphism of rings with kernel K. Then f is injective iff K = (0).

Let S be the set of all polynomials with degree less than n together with the zero polynomial. Then every congruence class modulo p(x) is the class of some polynomial in S and the congruence classes of different polynomials in S are distinct. Proof. 5 for the integers, the division algorithm says f (x) = q(x)p(x) + r(x) with deg r < deg p or r = 0. So r(x) ∈ S and f (x) ≡ r(x) (mod p(x)). Different polynomials in S have a difference of degree less than that of p, so the difference cannot be divisible by p, and therefore the polynomials are in distinct congruence classes.

### Abstract algebra [Lecture notes] by Thomas C. Craven

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