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On page 117 of (Corollary 4.6.12),

I'm not sure how/why the proof says that there exists "a unique subgroup of order 5." How did they come up with that conclusion?

Thanks in advance

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up vote 6 down vote accepted

Next time, please, quote the text.

So you want to know why the group $\mathbf{Z}_{11}^{*}$ has a unique subgroup of order $5$.

It is well known that $\mathbf{Z}_{11}^{*} = \mathbf{Z}_{11} \setminus \{ 0 \}$ is cyclic, of order $10$. Now a cyclic group $G$ of order $n$ has exactly one subgroup $H$ of order $m$ for each divisor $m$ of $n$. If $G = \langle a \rangle$, then $H = \langle a^{n/m} \rangle$.

In this particular case $\mathbf{Z}_{11}^{*} = \langle \bar{2} \rangle$, so the unique subgroup of order $5$ is $\langle \bar{2}^{2} \rangle = \langle \bar{4} \rangle$.

PS Perhaps it might be mentioned that any group of order $10$ (there are two isomorphism classes, cyclic and dihedral) has a unique subgroup of order $5$. It has one such subgroup by Sylow's first theorem (or Cauchy's theorem), and this is normal, because it has index $2$, and thus unique, by Sylow's second theorem. (Thanks a bunch to Marc van Leeuwen for pointing out an ambiguity in a previous version.)

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Just a caveat "normal, and thus unique" might suggest something horribly wrong to the naive reader. Sylow subgroups that happen to be normal are the unique subgroups of their order, because all Sylow subgroups are conjugated to each other. – Marc van Leeuwen Mar 18 '13 at 11:27
@MarcvanLeeuwen, ok, I will edit to avoid possible misunderstandings. Thanks! – Andreas Caranti Mar 18 '13 at 11:29
Of course, a cyclic group of order $10$ is small enough that someone who doesn't know any of these theorems could just inspect the group, find all elements of order $5$, and observe that they all generate the same cyclic subgroup. – Andreas Blass Mar 18 '13 at 13:17

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