I am currently reading http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/group12.pdf, and have a quick question about the group being non-abelian. Let me explain:

Let $|G|=12=2^2\cdot 3$ and let $n_2,$ $n_3$ denote the number of Sylow $2,3$-subgroups respectively. By using Sylow's Theorems and counting elements we see that either $n_3=1$, or $n_3=4$ (and $n_2=1$). So we always have a normal Sylow $2$-subgroup or a normal Sylow $3$-subgroup.

On page 2 the author turns to the case $n_2\ne 1$ (so $n_2=3$, $n_3=1$) and says "Since $n_2\ne 1$, the group is non-abelian". Why is this true?


By the second Sylow theorem, any Sylow 2-subgroups are conjugates of each other. But in an abelian group, every conjugate is the same subgroup again.

  • $\begingroup$ So in an abelian group any $n_p$ (denoting the number of Sylow $p$-subgroups) must be $1$? $\endgroup$ – Phil-ZXX May 12 '13 at 22:47
  • 1
    $\begingroup$ @Thomas Yes. Any finite abelian group is a direct product of its Sylow subgroups. $\endgroup$ – Joe Johnson 126 May 12 '13 at 22:48
  • $\begingroup$ @JoeJohnson126 So e.g. $C_3\times C_4\times C_4$ is the product of its Sylow 3-subgroup $C_3\times \{e\}\times \{e\}$ and its Sylow $2$-subgroup $\{e\}\times C_4\times C_4$ ? $\endgroup$ – Phil-ZXX May 12 '13 at 22:54

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