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Show that every subgroup of $Q_8$ is normal.

Is there any sophisticated way to do this ? I mean without needing to calculate everything out.

Alexander Gruber
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Kasper
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  • A better statement (or fact) is "if $G$ is a non-abelian $2$-group in which, every subgroup is normal, then it is (generalised) quaternion: $Q_{2^n}=\langle x,y\colon x^{2^{n-1}}, y^2=x^{2^{n-2}}, y^{-1}xy=x^{-1}\rangle$, $n\geq 3$". – Beginner Apr 17 '13 at 03:46

2 Answers2

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Here, I consider $$Q_8=\langle -1,i,j,k\mid (-1)^2=1,i^2=j^2=k^2=ijk=-1\rangle.$$ Note that $-1$ commutes with everything, and that all other non-identity elements have order $4$, so their cyclic subgroups have index $2$, and are therefore normal subgroups. Trivial subgroups are always normal, so since the non-trivial subgroups of $Q_8$ are $\langle -1\rangle,$ $\langle i\rangle,$ $\langle j\rangle,$ and $\langle k\rangle,$ then we're done.

Community
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Cameron Buie
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It depends what you call sophisticated. There is only one subgroup of order $8,$ one subgroup of order $2$ and one subgroup of order $1$ in $Q_{8},$ so each of those is normal. In any finite $p$-group, every maximal subgroup is normal, so each subgroup of order $4$ of $Q_{8}$ is normal. This is not really substantially different from Cameron Buie's more explicit answer.

Geoff Robinson
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