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I am having trouble with this very small proof and it is driving me crazy. I was hoping someone could give me some help:

If $ G $ is a noncyclic group of order $ 4p $ for some odd prime $ p \geq 5 $, then $ G $ contains an element of order 2.

Thanks.

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It would be helpful if you could say something about what theorems you already know. The assumptions are a bit odd, as the statement is true (and follows from a standard theorem) also for cyclic groups or if p < 5. This makes me wonder whether there's something specific about these kinds of groups which you've proved in class or earlier in the problem set which you could use. Alternately, if this is the first part of several questions about this group then maybe you're just expected to apply the standard theorem. –  Noah Snyder Nov 2 '12 at 22:54
    
We are supposed to use the following theorem: Let G be a group, where $ r \in G $, and suppose that $s \in G \notin \langle r \rangle$. If $ \langle r \rangle \cap \langle s \rangle = e $ and the order of $ s $ is $ k $, then $\langle r \rangle$, $s\langle r \rangle$,...,$s^{k-1}\langle r \rangle$ are distinct left cosets. –  mkeachie Nov 2 '12 at 23:04
    
Sorry there is a typo in there: $ s \in G \setminus \langle r \rangle$. –  mkeachie Nov 2 '12 at 23:06
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@dado, the converse of Lagrange's Theorem is not true. –  mkeachie Nov 2 '12 at 23:10

3 Answers 3

This is to me a very strange formulation of the problem. Any group $G$ with $|G|$ even has an element of order $2$. No need for $G$ to be noncyclic or the other condition. This is a special case of Cauchy's Theorem, but for the case at hand a slightly simpler proof than (any one of) the standard proofs of Cauchy's Theorem is given as follows.

Consider the action of the group ${1,-1}$ on $G$ given by $ag=g^a$. Notice that an element $g\in G$ has orbit size equal to $1$ iff $g^2=e$. Since the orbits partition $G$ and each orbit has size either $1$ or $2$, and $|G|$ being even, if there is one orbit of size $1$ there has to be another one. Since the orbit of $e$ has size $1$ it follows there is some $g\in G$ whose orbit has size is $1$, thus $g^2=e$ and an element of order $2$ is found.

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Suppose the finite group $G$ has an even number of elements. (Note: clearly $4p$ is even.)

Imagine a Cayley table (a group multiplication table).

Suppose for the sake of contradiction that no element has order $2$.

Then there is only one $e$ along the NW/SE diagonal of the table; namely, for $e*e = e$.

But we know that $xy = e$ implies $yx = e$ as well, so every other $e$ in the table comes in a pair symmetric around the aforementioned diagonal.

Therefore, the number of $e$'s in the table is odd: the one in the diagonal and the rest in pairs.

But the number of $e$'s in the table is also even: one in each row, and $|G|$ is even.

Contradiction.

Thus, we were wrong in our supposition, and there must be an element of order $2$.

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If the group has any elements whose order is a multiple of 2, then you're done. So the only way you could run into trouble is if every nontrivial element has order p. Now apply the theorem you mention in comments.

To my mind this is a slightly odd way of ending the argument. If every nontrivial element has order p, then you have 4p-1 nontrivial elements grouped into subgroups each with p-1 nontrivial elements and no overlaps. But then $(p-1) \mid (4p-1)$, so $(p-1) \mid 3$, which is impossible because p is odd.

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When you apply the theorem, do we let $ r=p $ and $s=e$ which leads to a contradiction? –  mkeachie Nov 2 '12 at 23:45
    
s=e isn't going to tell you anything, you have to take both r and s be nontrivial. –  Noah Snyder Nov 3 '12 at 0:09
    
Oops that is what I meant. –  mkeachie Nov 3 '12 at 0:11
    
In this hypothetical group of 4p where every nontrivial element has order p, how would you select two nontrivial elements to ensure that the subgroups they generate only share the identity element in common (so we can apply the theorem)? –  mkeachie Nov 3 '12 at 0:24

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