p is a prime number.
Let G be the Group of order $p^3$ whose every element in $G$ has order p.
Then G is not isomorphic to any subgroup of $GL(2; \mathbb{C})$.
This claim is correcet? Please help me.
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1The identity element in $G$ cannot have order $p$. So every element different from the identity having order $p$. – Dietrich Burde Feb 26 '17 at 16:32
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For odd $p$ there are two isomorphism classes of groups in which every nonidentity element has order $p$,one abelian and one nonabelian. Are you trying to prove this for both of these types of groups or only for one of them and,if so, which? – Derek Holt Feb 26 '17 at 20:47
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Since I had no idea, I have noticed that thing. I will consider both types. – log log Feb 27 '17 at 11:48
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For $p=2$ there are exactly two non-abelian groups of order $p^3$, namely $D_4$ and $Q_8$, but both do not satisfy the assumption that all non-identity elements have order $p$. It remains the abelian group $(\mathbb{Z}/2)^3$. For $p>2$ the classification given in Theorem $4$ shows, there are only two groups of order $p^3$ where all non-identity elements have order $p$, namely $(\mathbb{Z}/p)^3$ and $Heis(\mathbb{Z}/(p))$. Now the finite subgroups of $GL(2,\mathbb{C})$ are all classified, and the list easily shows that neither $(\mathbb{Z}/p)^3$ nor $Heis(\mathbb{Z}/(p))$ for any prime $p$ is a subgroup of $GL(2,\mathbb{C})$.
Dietrich Burde
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I didn't downvote. But at the time I made my comment, you had written that for $p=2$ there are only two groups of order $p^3$, and you have now edited it. I will delete my earlier comment and upvote!!! – Derek Holt Feb 26 '17 at 21:50
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@DerekHolt Thank you. Yes, you are right, for $p=2$ I forgot to consider $C_{p^3}$. – Dietrich Burde Feb 26 '17 at 22:42
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(1) $|G|=p^3$ and $G$ is isomorphic to subgroup of $GL({\bf 2},\mathbb{C})$.
(2) This means that there is a faithful two dimensional representation of $G$ over $\mathbb{C}$.
(3) If this representation is not irreducible, then it means, image of $G$ essenntially sits as block diagonal matrices $\begin{bmatrix} * & 0 \\ 0 & * \end{bmatrix}$.
(4) But such matrices can at the most form a group of order $p^2$ (i.e. such subgroup, being image of a $p$-group, has order $1$, $p$ or $p^2$). So $\rho$ will not be faithful ($|G|=p^3$ but order of image is not $p^3$), contradiction
(5) So $\rho$ should be irreducible.
(6) It is standard result in representation theory that the dimension of an irreducible representation should be a divisor of order of group. So here, $2$ should divide $p^3$.
(7) Conclusion: $p$ must be $2$. Follow arguments of Dietrich Brude, and try to write details.
p Groups
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Nice! I was gonna type up an argument using the normalizer of the diagonally embedded $C_p\times C_p$ inside $GL_2$ to reach the conclusion that we must have $p=2$. Representation theory FTW. – Jyrki Lahtonen Feb 27 '17 at 07:20