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Consider the group of order 28 with Sylow $2$-subgroups that are cyclic. We can derive that the Sylow $7$-subgroup is normal, and that this group is uniquely determined by the relation $bab^{-1}=a^6$ where $a$ is a generator of a Sylow $7$-subgroup and $b$ is the generator of a Sylow $2$-subgroup.

Suppose I wish to find how many elements of each order there are, and the class equation of the group. I've tried a few different things, but have simply confused myself. The constraint is that I not use the orbit-stabilizer theorem or any techniques that are more sophisticated than the fact that the size of the conjugacy class of an element is the index of its centralizer. (Of course the Sylow theorems are fair game.) Any help would be appreciated.

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  • $\begingroup$ Of course I know that the Sylow 7-subgroup has 6 elements of order 7, as it is isomorphic to $\mathbb{Z}_7$. $\endgroup$ – Jonah Apr 15 '14 at 17:39
  • $\begingroup$ I think you just compute the centralizers of all elements. The style of elements are $a^ib^{\pm1}$, $a^0b^2$, $a^jb^2$, $a^0b^0$, $a^jb^0$ with $j \neq 0$. $\endgroup$ – Jack Schmidt Apr 15 '14 at 17:55
  • $\begingroup$ How would I go about computing the centralizers? And how does that help with finding how many elements of each order there are? $\endgroup$ – Jonah Apr 15 '14 at 17:59
  • $\begingroup$ See math.stackexchange.com/questions/236262/…. $\endgroup$ – Dietrich Burde Apr 15 '14 at 18:08
  • $\begingroup$ I did see that topic, but it did not address the issues I am having with the question. $\endgroup$ – Jonah Apr 15 '14 at 18:09
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Notice that every element of $G$ can be written uniquely in the form $a^i b^j$ for $i\in \{0,1,\ldots,7-1\}$ and $j\in \{0,2,\pm 1\}$. Notice that $b^2$ centralizes both $a$ and $b$.

The subgroup $\langle ab^2 \rangle = \{ ab^2, a^2b^0, a^3b^2, a^4b^0, a^5b^2, a^6b^0, a^0b^2, a^1b^0, a^2 b^2, a^3 b^0, a^4 b^2, a^5 b^0, a^6 b^2, a^0 b^0 \}$ is cyclic of order $2(7)$. It must be normal since it is index 2. The centralizers depend on whether there is an $a$ present: for $a^0b^0$ and $a^0b^2$ the centralizer is the whole group, but otherwise it is just this cyclic group of order $2(7)$.

We (hopefully?) know how to count elements of cyclic groups by order, so we only need to worry about the elements of $G$ outside of this cyclic group.

However, we calculate the centralizer of $b$: it is just $\langle b \rangle$ of index 7, so $b$ and $b^{-1}$ each have 7 conjugates, none of which are contained in $\langle ab^2 \rangle$. Hence we get:

  • $\{a^0 b^0\}$ - order 1, class size 1, 1 class
  • $\{a^0 b^2\}$ - order 2, class size 1, 1 class
  • $\{a^i b^0, a^{-i} b^0 \}$ - order 7, class size 2, $(7-1)/2=3$ classes
  • $\{a^i b^2, a^{-i} b^2 \}$ - order 14, class size 2,$(7-1)/2=3$ classes
  • $\{a^i b^{+1} : i = 0,1,2,\ldots,7-1\}$ - order 2, class size 7, 1 class
  • $\{a^i b^{-1} : i = 0,1,2,\ldots,7-1\}$ - order 2, class size 7, 1 class

This works for any odd prime $p$ replacing $7$, with $bab^{-1}=a^{p-1}$.

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There are exactly two nonabelian groups of order $28$, namely the dihedral group $D_{14}$ and another group, namely a generalised quaternion group, which is a semidirect product of $C_7$ and $V_4$. A proof can be found along the lines in the comments here: Nonabelian order 28 group whose Sylow 2-subgroups are cyclic. By the way, not every group of order $28$ has cyclic $2$-Sylow subgroups, e.g., not $D_{14}$. So we are looking for the second group $C_7\rtimes V_4$.
Complete solutions to this popular homework question can be found at many places: see section $7$ in http://www.math.drexel.edu/~rboyer/courses/math533_03/sylow_thm.pdf, and section $4$ in http://math.arizona.edu/~savitt/teaching/07math511a/sol12.pdf.

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