Classification of groups of order 8

[mathusers]

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

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

(i) show that $H \lhd G$ and that $x^2 \epsilon H$
(ii) list, a priori, the possibilities for $x^2$ and label them $P_1,...,P_n$.
(iii) list the possibilities for $xyx^{-1}$ (HINT: CHECK ORDERS) and label them $Q_1,...,Q_m$
(iv) By examining the pairs, $(P_i,Q_j)$ in turn show that G is isomorphic to one of $C_8, C_4 \times C_2, D_8, Q_8$.
Deduce that an arbitrary group of order 8 is isomorphic to one of $C_2 \times C_2 \times C_2, C_4 \times C_2, C_8, D_8, Q_8$
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For part (i) i understand i have to show G is a normal subgroup of H but I am not sure how to show that G is a subgroup (i.e. closure holds, existence of identity element of G in H, etc) and it is normal. i..e For all g in G, gHg^−1 ⊆ N.

 

[Mathdope]

For part (i) i understand i have to show G is a normal subgroup of H but I am not sure how to show that G is a subgroup (i.e. closure holds, existence of identity element of G in H, etc) and it is normal. i..e For all g in G, gHg^−1 ⊆ N.

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?