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Does there exist a vector space with 30 elements? How to determine whether there exist any vector space of particular cardinality?

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Relevant: – anon May 8 '12 at 10:30
very useful thanks – srijan May 8 '12 at 10:50
up vote 10 down vote accepted

Any finite field has order $p^k$ for some prime $p$ and $k\geq1$. Then a vector space of dimension $n$ over such a field has $(p^k)^n=p^{kn}$ elements, so in particular, the number of elements of a vector space over a finite field must be a prime power. So there is no vector space with $30$ elements.

So for a general integer $x$, if $x\ne p^k$ for some $k\geq1$ and prime $p$ then there is no vector space of order $x$, and if $x=p^k$ for such $p$ and $k$, there is one vector space up to isomorphism for every (ordered) factorization of $k$ into two integers; if $k=k_1k_2$ then the $k_2$ dimensional vector space over the field of $p^{k_1}$ elements has order $x$.

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i got my answer.thank you sir – srijan May 8 '12 at 10:28
Dear sir what about its sub spaces? can it have subspaces having dimension lying between 0 to 30? – srijan May 8 '12 at 10:44
Well, a subspace is also a vector space, so there are the same conditions on possible cardinalities (I assume you meant cardinality, not dimension here). The $k_2$ dimensional vector space over the field of $p^{k_1}$ elements has subspaces of cardinality $p^{k_1t}$ for $0\leq t\leq k_2$. – Matthew Pressland May 8 '12 at 10:47
Thanks sir everything is cleared now – srijan May 8 '12 at 10:53

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