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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.2012-11-02
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    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.2012-11-02
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    Sorry there is a typo in there: $ s \in G \setminus \langle r \rangle$.2012-11-02
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    @dado, the converse of Lagrange's Theorem is not true.2012-11-02

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