# 3-Characterizations of finite groups by Makhnev A. A.

By Makhnev A. A.

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It remains to show that H ~ H x. Choose any h E H, then h = (hx-I)x E H x because h, x-I E H. Thus H ~ H x and so H = H x. (ii) Notice that if Hx = Hy, then Hxy-I = Hyy-l so Hxy-l = H. On the other hand, if Hxy-l = H, then Hxy-ly = Hy so Hx = Hy. Now apply part (i). 21 1. The Elements (iii) If 9 E Hx, then 9 = hx for some h E H, so Hg = Hhx = Hx. Therefore Hx ~ {g I Hx = Hg}. Conversely if Hx = Hg, then 9 = 1· 9 E Hg = Hx so {g I Hx = Hg} ~ Hx. We are done. (iv) This is a triviality, since x = 1 .

Proof Suppose that there are i, j E Z with i < j and xi = x j (if there are no such i and j we are in case (ii)). Among all such i and j choose a pair with j - i = n as small as possible. Now Xi = x j so 1 = X-iX i = x-ix j = xn. Moreover, Xo, Xl, ... , x n - 1 must be distinct otherwise we would violate the choice of n. Finally, if a E Z we can find q, r E Z with 0 ~ r < n such that a = qn + r. Now x a = x qn+T = (xn)qx T = lqx T = x T • Thus G = {XO,xI, ... ,xn- 1} and case (i) is established.

Choose any h E H, then h = (hx-I)x E H x because h, x-I E H. Thus H ~ H x and so H = H x. (ii) Notice that if Hx = Hy, then Hxy-I = Hyy-l so Hxy-l = H. On the other hand, if Hxy-l = H, then Hxy-ly = Hy so Hx = Hy. Now apply part (i). 21 1. The Elements (iii) If 9 E Hx, then 9 = hx for some h E H, so Hg = Hhx = Hx. Therefore Hx ~ {g I Hx = Hg}. Conversely if Hx = Hg, then 9 = 1· 9 E Hg = Hx so {g I Hx = Hg} ~ Hx. We are done. (iv) This is a triviality, since x = 1 . x E H x. 18 gives a criterion for deciding when two (right) cosets of H in G are equal.

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