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Let $a,b$ be elements in a group $G$ such that $|a|=25$ and $|b|=49$. Prove that $G$ contains an element of order 35.

If $G$ is finite then we can say that $5,7 \mid G$ and hence by Cauchy's Theorem, we have elements $c,d$ in $G$ of orders 5 and 7, respectively. So we can take $cd$ to be the element of order 35. What about the case when $G$ is infinite?

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"So we can take $cd$ to be the element of order 35." Is $G$ abelian? – anon Apr 4 '12 at 5:09
@anon No. Oops, so we can't say that. – Galois Apr 4 '12 at 5:12
And in any case, you already have elements of order $5$ and of order $7$ (whether $G$ is finite or note), namely $a^5$ and $b^7$, respectively. – Arturo Magidin Apr 4 '12 at 5:18
Where does this claim come from? – user641 Apr 4 '12 at 6:12
There are finite counterexamples as well. The companion matrix of the 49th cyclotomic polynomial (of degree 42) over Z/25Z has no eigenvalues that are 7th roots of unity, and in particular, its 7th power acts fixed point freely on (Z/25Z)^42, and the semi-direct product has no elements of order 35, but elements of orders 25 and 49. I think reversing the roles of 5 and 7 gives a smaller group, but the argument is a little more complicated as the 25th cyclotomic polynomial is reducible mod 49. – Jack Schmidt Apr 4 '12 at 15:47
up vote 13 down vote accepted

As stated, the claim is false: let $G=C_{25}*C_{49}$, the free product of a cyclic group of order $25$ and a cyclic group of order $49$; if $a$ is a generator of $C_{25}$ and $b$ is a generator of $C_{49}$, then $a$ and $b$ have the desired properties.

However, in a free product $H*K$, the only elements of finite order are conjugate to elements of finite order in $H$ or in $K$; so in $C_{25}*C_{49}$, the only elements of finite order are conjugate to an element of $C_{25}$ or to an element of $C_{49}$, and thus must have order $1$, $5$, $25$, $7$, or $49$; no element of $G$ has order $35$.

So presumably, there are some extra hypothesis on $G$ to make the implication true...

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Such as commutativity.... – Gerry Myerson Apr 4 '12 at 5:55
@Gerry: Sure, that would be easy, and the OP already has the answer then. (-: – Arturo Magidin Apr 4 '12 at 5:57
@Gerry, or G finite and nilpotent ... – Nicky Hekster Apr 4 '12 at 8:47

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