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Classification of groups of order 8

  1. Feb 27, 2008 #1
    Hi next one, bit confused with this problem: any hints on any of the parts would be greatly appreciated.

    QUESTION:
    ---------------------------------------
    let G be a group of order 8 and suppose that [itex]y \epsilon G[/itex] has ord(y)=4. Put [itex]H = [1,y,y^2,y^3][/itex] and let [itex]x \epsilon G-H[/itex]

    (i) show that [itex]H \lhd G[/itex] and that [itex]x^2 \epsilon H[/itex]
    (ii) list, a priori, the possibilities for [itex]x^2[/itex] and label them [itex]P_1,...,P_n[/itex].
    (iii) list the possibilities for [itex]xyx^{-1}[/itex] (HINT: CHECK ORDERS) and label them [itex]Q_1,....,Q_m[/itex]
    (iv) By examining the pairs, [itex](P_i,Q_j)[/itex] in turn show that G is isomorphic to one of [itex]C_8, C_4 \times C_2, D_8, Q_8[/itex].
    Deduce that an arbitrary group of order 8 is isomorphic to one of [itex]C_2 \times C_2 \times C_2, C_4 \times C_2, C_8, D_8, Q_8[/itex]
    ---------------------------------------

    For part (i) i understand i have to show G is a normal subgroup of H but im not sure how to show that G is a subgroup (i.e. closure holds, existance of identity element of G in H, etc) and it is normal. i..e For all g in G, gHg^−1 ⊆ N.
     
  2. jcsd
  3. Feb 27, 2008 #2
    You want to show H is a normal subgroup of G not the reverse. Showing that H is a subgroup follows from the fact that y has order 4. Do you see why? To show it's a normal subgroup of G consider the left (or right) cosets of H in G. How many are there?
     
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