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I got stuck on this for a bit, then came across Cauchy's theorem:

Let G be a finite group and let p be a prime that divides the order of G. Then G has an element of order p.

G is obviously finite, and 2 is a prime that divides the group. Is this all I need to prove it? I feel that was too easy and am probably missing something. Mind you, we have yet to get to that chapter so maybe "technically" I can't use it, but is it still valid to do so?

Thanks, Jeff

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  • $\begingroup$ Yes it is valid to do that, but it also makes it basically trivial, so you're probably right in thinking you are expected to do the question without simply stating the theorem. $\endgroup$ – John Doe Jun 26 '18 at 17:03
  • $\begingroup$ @John Doe. Thanks, it's nice to know that if nothing else, I won't leave a blank answer! Any hints on how I can solve it otherwise? I know the order must divide the group, so we have possible values of { 1, 2, 4, 11, 22, 44}. But it can't be the identity, so that eliminates 1. I saw a similar question (for order 8), and they made the assumption that if G was cyclic and had a generator <a>, where a^8 = e , this implies that there exists an element a^4 whose order is 2. Is this true? If so, how so? I don't see the connection. Thanks. $\endgroup$ – Jeff Jun 26 '18 at 17:25
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Hint:

For an elementary proof, just factor $44$ and note it has only $6$ divisors. Then Lagrange's theorem tells you the order of an element is one of these divisors. Examine each case, observe that if there's no element of order $2$, there's no element of order an even number either, so all elements $\ne e$ have order $11$. Is it compatible with $|C|=44$?

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  • $\begingroup$ thanks. That's where I'm having my struggle, is how to properly examine each one of the divisors. I could use exponents rules and get them all to equal the identity, but that wouldn't do any good right? Because it can't be the identity I don't think. $\endgroup$ – Jeff Jun 26 '18 at 17:42
  • $\begingroup$ Do you agree that if there's an element of even order, there's an element of order $2$? $\endgroup$ – Bernard Jun 26 '18 at 17:46
  • $\begingroup$ So if we have a group G = {x, y, z, t}. Then |G| = 4. So it must be that x, y, z or t in G must have order 2? Sounds true. But I can't prove it I don't think. $\endgroup$ – Jeff Jun 26 '18 at 18:10
  • $\begingroup$ Either it's not cycllic, and $3$ elements have order a strict divisor of $4$, i.e. $2$. Or it's cyclic, and there's a element, say $x$, with order $4$. What's the order of $x^2$? $\endgroup$ – Bernard Jun 26 '18 at 18:44
  • $\begingroup$ Some details on Bernard's nice argument (the way I understand it): Assume there is no element of order 2. Then all nontrivial elements have order 11. Let $H_1,...,H_n$ be the distinct subgroups of order 11. For any group $G$ set $G':=G\setminus\{1\}$. For $1\le i<j\le n$ we have $H_i\cap H_j=\{1\}$. Thus $C'$ is the disjoint union of the $H_i'$, which gives $43=10n$. $\endgroup$ – Pierre-Yves Gaillard Jun 26 '18 at 23:24
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Cauchy's theorem for $p=2$ has a simple elementary proof.

Let $G$ be a finite group of even order. Then $G$ has an element of order $2$.

The orbits of the map $x \mapsto x^{-1}$ partition $G$ into the subsets $\mathcal O(x)=\{x, x^{-1}\}$. These subsets have cardinality $1$ or $2$. In fact, $\mathcal O(x)$ has one cardinality $1$ iff $x^2=1$ iff $x=e$ or $x$ has order $2$.

Let $m_k$ be the number of such subsets that have cardinality $k$. If $G$ has order $n$, then $n=m_1+2m_2$. Since $n$ is even, so is $m_1$. Now, $m_1\ge1$ because of $|\mathcal O(e)|=1$. Therefore, $m_1 \ge 2$. In other words, there is an element of order $2$.

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