Let $G$ be a non abelian group; then its order can be:

  1. $25$

  2. $55$

  3. $35$

  4. $125$

I think the order cannot be $25$ and $35$. But from option $55$ and $125$ which one is not possible?

Why not $25$ because every group of order $p^2$ is abelian. Here $p$ is $5$. Why not $35$ because $35=5*7$ and $5$ does not divide $7-1$. There there is only one group which is cyclic up to isomorphism of order $35$.

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    $\begingroup$ you tried something?? $\endgroup$ – user87543 Jan 30 '15 at 10:26
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    $\begingroup$ $55$ and $125$ are both possible. $\endgroup$ – Ofir Schnabel Jan 30 '15 at 10:30
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    $\begingroup$ First of all mind your language.... I really do not see any thing other than "But from option 55 and 125 which one is not possible?" You did not said anything about what do you think about those groups.... $\endgroup$ – user87543 Jan 30 '15 at 10:30
  • $\begingroup$ @ketan why you think $|G|\neq 35$? what is the difference with $|G|=55$? $\endgroup$ – Ofir Schnabel Jan 30 '15 at 10:41
  1. $|G|=25$ is not possible since groups of order $p^2$ are always abelian.

  2. $|G|=125$ is possible since for any prime $p$ there exist a non-abelian group of order $p^3$.

  3. In a group of order $35$ the $7$ group and the $5$ group are normal by the Sylow theorems and hence this group is cyclic.

  4. In a group of order $55$, the $11$-group is normal, but the $5$-group does not have to be normal and therefore there is a non-commutative group of order $55$.

Other words for $35,55$. In both cases the group must be a semi-direct product $$C_q\rtimes C_p,$$ where $p<q$. In the case $q=7$ this action must be trivial since $5$ do not divide $7-1$. However, for $q=11$ there is a non-trivial action of $C_5$ on $C_{11}$ and therefore the group do not have to be commutative.


Hint: 1)Note that $o(G)=5^2$,what do u know about groups of order $p^2$.

2)Note that $o(G)=5*11 $ what do u know about groups of order $pq$.

3)Note that $o(G)=5*7$ again what do u know about groups of order $pq$.

4)Note that $o(G)=5^3$,what do u know about groups of order $p^3$.

Added:1)Every group of order $p^2$ is abelian.

2,3)Suppose $o(G)=pq$, $p<q$ and p does not divide $q-1$ then every group of order $pq$ is cyclic.Suppose p divides $q-1$ then there is a unique nonabelian group of order $pq$,obviously upto isomorphism.

4)For any prime there exist a nonabelian group of order $p^3$,namely Heisenberg Group with entries from a field of order $p$.

  • $\begingroup$ What do we know about groups of order $pq$? Say on groups of order $6$? $\endgroup$ – Ofir Schnabel Jan 30 '15 at 10:38
  • $\begingroup$ obviously,there are at most two groups of order pq depending on p and q $\endgroup$ – Arpit Kansal Jan 30 '15 at 10:40
  • $\begingroup$ @OfirSchnabel did u downvote? why someone downvote my answer? $\endgroup$ – Arpit Kansal Jan 30 '15 at 10:50
  • $\begingroup$ I did because it was wrong and also you gave it before the OP show any effort of his own. Now your answer is good so I cancel my downvote. $\endgroup$ – Ofir Schnabel Jan 30 '15 at 10:57
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    $\begingroup$ From your original answer it looked like $35$ and $55$ are the same, you gave the same resoning. $\endgroup$ – Ofir Schnabel Jan 30 '15 at 11:00

Take G=HF(5), Heisenberg group over feild of order 5 gives non Abelian group of order 125


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