Show that any group of order 10 is either cyclic (so isomorphic toZ10),

or is isomorphic to the dihedral group D5 of order 10 (symmetries of a regular

pentagon).

I have no idea how to even get started on this

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- Mar 14th 2011, 12:30 PMpoirottricky group proofShow that any group of order 10 is either cyclic (so isomorphic toZ10),

or is isomorphic to the dihedral group D5 of order 10 (symmetries of a regular

pentagon).

I have no idea how to even get started on this

- Mar 14th 2011, 12:49 PMTheChaz
Just a thought from the logical side of things. You have:

$\displaystyle |G| = 10 \Rightarrow (G \cong \mathbb{Z}_{10} \vee G \cong D_5)$

So we can start by assuming that G is NOT isomorphic to Z10. This is a common approach to proofs of the form P -> (Q v R), where "v" is the logical operator "or".

In other words, not being Z10 means no elements have order ten. But the order of an element must divide the order of the group, so look for elements of orders 5 and 2. - Mar 14th 2011, 01:46 PMTinyboss
Have you studied Sylow's theorems, semidirect products, and the fundamental theorem of finitely-generated abelian groups yet? Those are the usual way to attack this kind of problem, although when the order is this small, you can usually make more basic arguments.

For instance, if the group is abelian it's cyclic, since 2 and 5 are coprime. You also know (in any case) that there's a normal subgroup of order 5 (Cauchy's theorem gives you the existence of an order-5 element, and the cyclic subgroup generated by it has index 2 in G, so it's normal). - Mar 14th 2011, 03:27 PMDrSteve
Let G be a group with 10 elements. The number of Sylow 5-subgroups must divide 10 (the possibilities are 1, 2, 5, 10) and be congruent to 1 modulo 5. Thus there is only one Sylow 5-subgroup, P, which in thus a normal subgroup of G. Since the order of P is 5, P is cyclic, say P = <x>. G also has at least one Sylow 2-subgroup, so let y be an element of order 2. Thus the elements of G are $\displaystyle \{e, x,x^2,x^3,x^4,y,yx,yx^2,yx^3,yx^4\}$. Since P is normal, $\displaystyle yxy^{-1}$ is in P, and so is of the form $\displaystyle x^i$ for some $\displaystyle i$.

If $\displaystyle i = -1$, then $\displaystyle yx = x^{-1}y$, and G is isomorphic to

$\displaystyle D(5) = <x, y | x^5 = e = y^2, yx = x^{-1}y>$.

Assume $\displaystyle i \ne -1$. Since $\displaystyle y^2 = e, (yx)^2 = yxy^{-1}x = x^{i+1}$. So the even powers of $\displaystyle yx$ are powers of x, whereas the odd powers of $\displaystyle yx$ are of the form $\displaystyle yx^j$ for some j. The order of $\displaystyle yx$ divides 10, and so is 1, 2, 5, or 10. We see that $\displaystyle yx$ is not of order 1 (otherwise $\displaystyle x = y$), it’s not of order 2 (since $\displaystyle (yx)^2 = x^{i+1}$), and it’s not of order 5 (since its 5th power is $\displaystyle yx^j$ for some j). Thus $\displaystyle yx$ has order 10, and G is cyclic (thus, G is abelian, and $\displaystyle yx = xy$). - Mar 14th 2011, 03:55 PMDrexel28
- Mar 15th 2011, 09:25 AMpoirot
No we haven't studied any of those theorems. Whats that result about abelian and co prime? Its irrelevant but just interested.

Anyway, I was thinking of proof along these lines.

By lagrange's theorem, elements (except identity) must have orders 2,5 or 10. If there is an element order 10, it is cyclic. If not such and such must be true. But I am not given any information on the elements.