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Assume that $G$ is a group such that $|G|=55$ and $G$ has exactly four elements with order $5$. Prove that center of $G$ is not trivial and $G$ is a cyclic group.

My try:
We know that $|h_1|=|h_2|=|h_3|=|h_4|=5$ and there does not exist $h_5$ such that $|h_5|=5$. Hence we know that $h_1^5=h_2^5=h_3^5=h_4^5=1$. If element has an order $1$ then it is neutral element in $G$ so $|e|=1$. Divisors of $G$ are $1,5,11,55$ so others $50$ elements which we haven't mentioned have an order $11$ or $55$.

Center of $G$ is a set: $Z(G)=\{g\in G: \forall _{x\in G} xg=gx\}$. The group is cyclic when is generated by one element.

These are all observations about this task. Have you got some ideas?

amWhy
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MP3129
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    Have you seen Sylow's theorem? – take008 Nov 23 '19 at 16:03
  • @take008 yes, I have seen: since $5$ and $11$ are prime numbers, from Sylow's thorem we have that: $$\exists _{H \subseteq G} |H|=5$$ $$\exists _{K \subseteq G} |K|=11$$ – MP3129 Nov 23 '19 at 16:17
  • Also by Sylow's theorem you have $K$ is characteristic, and by your given data $H$ is also characteristic. – take008 Nov 23 '19 at 16:22

3 Answers3

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Here is a hint:

Let $P_5$ and $P_{11}$ be sylow subgroups of $G$. Then $P_5 \cong \mathbb{Z}/5$ and $P_{11} \cong \mathbb{Z}/11$ since the only groups of prime order are cyclic.

Now if we can show $G = P_5 \times P_{11}$ then we are done (why?). So it suffices to show both are normal, since their intersection must be trivial (why?).

$P_{11}$ is normal as a consequence of Sylow 3, this is routine and I'll leave it to you.

$P_5$ is normal because we know there are exactly 4 elements of order 5, so $P_5$, which already has 4 elements of order 5, must be the only Sylow 5-subgroup. We can thus conclude that it is normal too (why?).

Have fun filling in the gaps! Feel free to comment if there's any follow up questions.

I hope this helps ^_^

HallaSurvivor
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As a corollary of the assumption, $P_5$ is unique, hence normal, hence union of conjugacy classes. If the center was trivial, then the union of conjugacy classes forming $P_5$ would read (in terms of sizes) $5=1+2+2$ or $5=1+4$, but $G$ hasn't got conjugacy classes of size $2$ or $4$, as it has odd order. Therefore $P_5$ is necessarily the union of five singleton classes, namely it is central. But since $G/P_5$ is cyclic, this implies that $G$ is abelian, hence cyclic (take two elements of order $11$ and $5$; their product has order $55$).

citadel
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The Sylow $5$ is unique because there's only $4$ elements of order $5$. Hence it's normal.

Then since the Sylow $5,11$ intersect trivially, we easily get that $G\cong\Bbb Z_5\rtimes \Bbb Z_{11}.$

But there's no non-trivial action because $(4,11)=1.$

So it's cyclic and the center is everything.

calc ll
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