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My question is about group theory:

How many subgroups does a non-cyclic group contain whose order is 25?

How can i answer that question?

Can you generalize the answer?

Thanks for your help.

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3 Answers 3

There are three possibilities for the order of any subgroup $H$ of a group $G$ of order $25$:

  • $|H| = 1 \iff H = e$
  • $|H| = 5,$ since $5\mid 25$.
  • $|H| = 25$, if $H = G$.

We're given that $G$ is non-cyclic, so the order of any element $x \neq e$ must be $5$, else, if $25$, it would generate the group, and hence the group would be cyclic. (Contradiction). I.e., $x \neq e \; \implies \;|\langle x \rangle| = 5$, and for each distinct subgroup $\langle x_i\rangle = \{e, x_i, x_i^2, x_i^3, x_i^4\}$, the elements $x_i, x_i^2, x_i^3, x_i^4$ are each of order $5$, and any one of them generates the same subgroup as does $x_i$.

Since $G$ is non cyclic, and $|G| = 25 = 5^2,$ where $5$ is prime, we know that $G \cong \mathbb Z_5 \times \mathbb Z_5$

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how can we find the number of the subgroups whose order is 5? –  juliet Jul 10 '13 at 15:27
    
Hmmm, there must be six distinct $x_i$, $1\leq i \leq 6$, each of which each generates a distinct subgroup. Since each distinct subgroup has 4 elements generating the same subgroup : this gives us $6\times 4$ plus the identity, for a total of $25$ elements in G. So we'd have $6$ subgroups of order $5$. –  amWhy Jul 10 '13 at 15:31
    
@amWhy: deserves another TU +1 –  Amzoti Jul 11 '13 at 1:21

As a power of a prime, a non-cyclic group $\;G\;$ of order $\;25\;$ is isomorphic to $\,C_5\times C_5\;$ , with $\,C_5\cong\Bbb Z/5\Bbb Z\;$ is the cyclic group of order $\;5\;$

Now, some hints for you to answer your question:

== A group of the form $\;G=\underbrace{C_p\times C_p\times\ldots\times C_p}_{n\;\text{times}}\;$ is a vector space of dimension $\;n\;$ over the field $\,\Bbb Z/p\Bbb Z\;$ and is therefore isomorphic to $\;\left(\Bbb Z/p\Bbb Z\right)^n\;$

== The subgroups of $\,G\,$ as above are in $\,1-1\,$ correspondence with the vector subspaces of $\,\left(\Bbb Z/p\Bbb Z\right)^n\;$

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I think that even though it is stated in the question, someone who falls directly on your answer would need to read "a non-cyclic group $G$ of order $25$...". Also, it is only true for $p^2$. Not any prime power. –  1015 Jul 10 '13 at 15:25
    
I think you're right, @julien: I'll add this. Thanks. –  DonAntonio Jul 10 '13 at 15:26

If $H$ is a non-trivial subgroup, then $H$ has $5$ elements (why?).

As the subgroup is not cyclic, and $25=5^2$, the order of every element $x \neq e$ is .....

Last but not least, if the order of $x$ is 5, there are 4 other elements of order 5 which generate the same group.

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