Let H be a subgroup of a finite group G. (a) Suppose that Ha = bH for two elements a, b ∈ G – H (in G, but not in H). Show that aH = Hb. (b) Suppose in addition that | G : H | = 3. Show that H is a normal subgroup of G.
Added by Carolyn M.
Step 1
This means that there exist h1, h2 ∈ H such that a = bh1 and b = ah2. Then we have: aH = bh1H = bHh1 = Hbh1 = Hah2 = Ha(h2H) = Ha where we have used the fact that H is a subgroup of G and hence is closed under multiplication and contains the identity element. Show more…
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