Show that none of the five subgroups of order $2$ in $D_5$ are normal.
Can I say $\alpha^2 H\neq H\alpha^2$ for $\alpha^2\in D_5$, with $\alpha^2$ having an order of $2$
Show that none of the five subgroups of order $2$ in $D_5$ are normal.
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$\begingroup$
abstract-algebra
group-theory
finite-groups
dihedral-groups
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1I guess that depends on what $H$ is. Also which $\alpha\in D_5$ gives an $\alpha^2$ of order $2$? – 2017-02-24
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0I thought it was the rotations but I think I am mistaken. – 2017-02-24
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2Hint. The rotations have order 5; the reflections have order 2. Work out the geometry of $ABA^-1$ when $A$ is a rotation and $B$ is a reflection. Or do this calculation expressing the group elements as permutations of the vertices. – 2017-02-24
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0As @EthanBolker notes in his comment on my answer, you should probably provide some more context so you get an appropriate-level answer. If you're happy to have some that use whatever is available, of course, feel free to leave things as-is. Cheers! – 2017-02-24
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1A normal subgroup of order 2 is central. The quotient would have order 5, so be cyclic. What do you know about groups who have cyclic central quotients? – 2017-02-24
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0The answer of Steve D is the most elegant one, a proof with hands down, without any appeal to Sylow Theory. +1 from me! – 2017-02-24
1 Answers
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Sylow's theorem indicates that if $P_2$ is normal it is unique. Since you have already stated $\#P_2>1$ that is sufficient.
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0Correct, of course - but I suspect the OP hasn't reached the Sylow theorems yet in his study of group theory. – 2017-02-24