If N is a normal subgroup of G and N is a cycle group, show that for any H (a su

Question

Explanation / Answer

Clearly NH is nonempty, because e ? N n H ==> e = e * e ? NH. -------------------------------------- Let n, n' ? N, and h, h' ? H. ==> nh and n'h' ? NH. (i) Closure under multiplication. Note that (nh)(n'h') = n (hn' h?¹) hh'. Since N is normal in G and H is a subgroup of G, we have (hn' h?¹) ? N. Hence, n (hn' h?¹) ? N, by closure in N. Moreover, hh' is in H, by closure of H. Hence, (nh)(n'h') = n (hn' h?¹) hh' ? NH. ------------------------ (ii) Inverses. Given nh ? NH, we need to show that (nh)?¹ = h?¹ n?¹ ? NH. To show this, note that h?¹ n?¹ = (h?¹ n?¹ h) h?¹. As before, (h?¹ n?¹ h) ? N, while h?¹ ? H. Hence, h?¹ n?¹ = (h?¹ n?¹ h) h?¹ ? NH. ---------------------- Therefore, NH is a subgroup of G. I hope this helps!

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.