By Randall R. Holmes

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**Additional resources for Algebra I**

**Sample text**

Theorem. If |G : H| = 2, then H G. Proof. Exercise 6–2. 7 Normality is not transitive Let G be a group. If H ≤ K ≤ G, then H ≤ G, so the property of being a subgroup is transitive. However, it is not true in general that if H K G, then H G, as the following example shows. • Let G be the dihedral group D8 of order 8. 3. One easily checks that ρ2 , and hence N , is in the center of G so that N is normal. 5) and it contains H. 3), so K has order 4. 6. 3). This example shows that, in general, normality is not transitive.

The proof is complete. 4(ii). The equivalence of (i) and (iii) suggests that it might be the case that, for any g ∈ G, we have the equality H g = H if and only if H g ⊆ H. This does not hold in general (see Exercise 6–1), but it does hold if G is finite. Let H G and let g ∈ G. In view of (i)⇒(ii) we can drop the terms “left” and “right” in referring to cosets of H and define the coset of H determined by g to be either gH or Hg. We record a few simple observations about normal subgroups: • G and {e} are both normal subgroups of G.

For one thing, this example shows that a group can have more than one composition series. But is also suggests that any two composition series of 55 a group have the same factors (except possibly for the order in which they occur). 5. 3 Zassenhaus butterfly lemma Let G be a group and let A1 , A2 , B1 , B2 be subgroups of G with A2 A1 and B2 B1 . 5, each of the products A11 = (A1 ∩ B1 )A2 A12 = (A1 ∩ B2 )A2 B11 = (A1 ∩ B1 )B2 B21 = (A2 ∩ B1 )B2 is a subgroup of G. Moreover, A12 A11 and B21 B11 , as is revealed in the proof of the next result.