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I have been recently I asked to construct non isomorphic groups of specific order. To do that I used the invariant that they must have isomorphic centers. $G_1\cong G_2 \Rightarrow Z(G_1) \cong Z(G_2)$, which proved to be handy.

For instance $3$ non isomoprhic groups of order $42$.

$$G_1 =\mathbb{Z}_{42}, G_2=\mathbb{Z_3} \times D_{14}, G_3=\mathbb{Z_7}\times D_6$$ where $D$ stand stand for Dihedral. $$Z(G_1)= \mathbb{Z_{42}}, Z(G_2)=\mathbb{Z_3}\times 1 , Z(G_3)= \mathbb{Z_7} \times 1$$ thus they are non isomorphic.

Now from sylow thereoms we could tell that the number of sylow subgroups of a Group of order $2$ could be $1,3,7,21$. So if for each different of these numbers a Group existed then we would have our 4 non isomorphic groups. Is a way to construct groups with the possible candidate numbers from sylow theorems?

If not is there something close to that end?

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  • $\begingroup$ Can I ask you to clarify? Are you saying there could be 1, 3, 7, 21 sylow subgroups of order 2 if your group is of order 42? or are you saying there could be sylow subgroups of order 1, 3, 7, and 21? The 1, 3 and 7 are certainly guaranteed if that is what you are saying. $\endgroup$
    – Betty Mock
    Commented Aug 31, 2013 at 3:08
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    $\begingroup$ The congruence condition from the Sylow theorems does not always guarantee existence: while $22 \equiv 1 \bmod 3$, no finite group has 22 3-Sylow subgroups. See M. Hall, On the Number of Sylow Subgroups in a Finite Group, J. Algebra 7 (1961), 363-371. $\endgroup$
    – KCd
    Commented Aug 31, 2013 at 3:58
  • $\begingroup$ @BettyMock The first one, that if the Group has order 42 then the number of $\, \mathrm{Sylow_2}\,$ subgroups is either one of $1,,3,7,21$. I dont think is the trivial subgroup is regarded a sylow subgroup though $\endgroup$
    – clark
    Commented Aug 31, 2013 at 4:07
  • $\begingroup$ @KCd I see so it does guarantee existence.. although it felt intuitive I didnt know where to seek a counterexample. Thanks for the reference! $\endgroup$
    – clark
    Commented Aug 31, 2013 at 4:11
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    $\begingroup$ @clark: (1) to fix your comment: "...so it does not guarantee existence." (2) The trivial subgroup is definitely considered to be a $p$-Sylow subgroup if $p$ doesn't divide the order of the group. There is nothing in the Sylow theorems that forces $p$-Sylow subgroups to be nontrivial; the proof goes through using the definition of Sylow subgroups as subgroups with maximal $p$-power order (which is order 1 if $p$ isn't a factor of the size of the group). $\endgroup$
    – KCd
    Commented Aug 31, 2013 at 13:49

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