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I am trying construct some examples of subsets of $S_{10}$ that are not subgroups? For instance would the set $\{\beta \in S_{10} : \beta(9) = 10\}$ not be a subgroup? Any examples with verifications would be great.

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fr56
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  • Keep in mind the fact that the order of a subgroup divides the order of a group. So, any set of elements that has a cardinality that is NOT a factor of 10! will do as well. (Specifically, sets whose cardinality have a prime factor of 11 or higher) – Alan Nov 15 '14 at 05:11
  • Also, anything that doesn't include the identity won't be a subgroup, trivially – Alan Nov 15 '14 at 05:12

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The set you constructed is not a subgroup, for if it was, it would have to contain $\beta^2$ for a given element $\beta$, but $\beta^2(9)=\beta(\beta(9))=\beta(10) \neq 10$ since $\beta$ is injective.

Nocturne
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    Great point! @Nocturne – fr56 Nov 15 '14 at 03:35
  • what if we say b^2=e? That is {b in S_10 | b^2 = e}? @Nocturne – fr56 Nov 15 '14 at 03:46
  • Again not a group. For it would contain, say $(12)$ and $(23)$ which are both of order 2, but would not contain their "product" $(12)(23)=(123)$ since $(123)$ has order 3. Note, however, that had your group been abelian, the set you constructed would have been a subgroup as the kernel of the homomorphism $\phi(x)=x^2$. – Nocturne Nov 15 '14 at 03:50
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The easiest way to do this is to take a set that isn't closed. For example, take any subgroup of $S_{10}$ that contains more than one element and remove one element.

Stella Biderman
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Any singleton subset (other than the one which is just the identity) suffices; the set $\{a\}$ isn't a group unless $a\cdot a = a$.

Milo Brandt
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