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In a cyclic subgroup of order $10$, there are $\phi(10)=4$ elements of order $10$.
Since there are exactly $8$ elements of order $10$, we can choose $4$ elements out of the $8$ elements of order $10$ in $^8C_4=70$ ways and for each way we have a cyclic subgroup of order $10$ ($4$ elements will of order 4 and the rest 4 will come from group G). Thus, $G$ has $70$ cyclic subgroups of order $10$.

But the answer to this question is $2$ since a cyclic subgroup of order $10$ can only have $4$ elements of order $10$ and hence there are $8/4$ cyclic subgroups only. But I don't understand why? There are $^8C_4=70$ ways to choose $4$ elements out of $8$ elements. So I think it should be $70$.

Please help me understand. Thanks in advance.

PS: I know that in a finite group, no. of subgroups should divide the order of the group. However, in my case since order of $G$ is not given, I am confused.

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  • $\begingroup$ Why do you suppose each of the $\binom{8}{4}$ ways of choosing four of the elements corresponds to a different subgroup? Let one of the eight elements be called $a$. Consider the subgroup generated by $a$... so you have $0, a, a+a, a+a+a, a+a+a+a, \dots, 9\cdot a$. Now... notice that $a, 3a, 7a,$ and $9a$ are all of order $10$. $\endgroup$ – JMoravitz Jun 1 '20 at 13:08
  • $\begingroup$ For building intuition, consider $\Bbb Z_{10}\times \Bbb Z_{10}$ $\endgroup$ – JMoravitz Jun 1 '20 at 13:08
  • $\begingroup$ @JMoravitz, I think I am starting to understand it now. If $a\in G$ is of order $10$ then $\{e,a,a^2,...,a^9\}$ is a subgroup of order 10 and it does have exactly $4$ elements of order 10 viz. $a,a^3,a^7,a^9$. Now $4$ are left to be utilized. $\endgroup$ – Koro Jun 1 '20 at 13:15
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    $\begingroup$ I don't think it is possible for a group to have exactly eight elements of order $10$, which would mean that the question was meaningless. $\endgroup$ – Derek Holt Jun 1 '20 at 13:35
  • $\begingroup$ @Derek Holt, it's an exercise problem from Gallian's abstract algebra. $\endgroup$ – Koro Jun 1 '20 at 13:39
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I claim that there is no group with exactly $8$ elements of order $10$, which makes the whole question logically meaningless (except possibly as a lemma in the proof that there is no such group).

Such a group $G$ would have two exactly two distinct subgroups $A$ and $B$ of order $10$. If they were not normal in $G$, then they would be conjugate and their normalizer would have index $2$ and contain both $A$ and $B$, so we could replace $G$ by this subgroup to get a smaller example.

So we can assume that $G = \langle A,B \rangle = AB$ with $A,B \unlhd G$.

Since $|{\rm Aut}(A)| = 4$, the element $g$ of order $5$ in $B$ must centralize $A$. Then, if $g \not\in A$, then $\langle g, A \rangle \cong C_5 \times C_{10}$ has $6$ subgroups of order $15$, contrary to assumption.

So $g \in A$, $|A \cap B| = 5$, and $|G| = |AB| = 20$.

Now, of the five (isomorphism types of) groups of order $20$, $C_{20}$, $D_{20}$ and the dicyclic group have a unique subgroup of order $10$, $C_2 \times C_{10}$ has three such subgroups, and the final group, which is a semidirect product $C_5 \rtimes C_4$ with faithful action has none.

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  • $\begingroup$ His question makes sense you've proven no such object exists. $\endgroup$ – cactus314 Jun 1 '20 at 20:09
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    $\begingroup$ @cactus314 Yes, but the statement "All groups that have exactly eight elements of order 10 have 263 subgroup of order 10" is true! $\endgroup$ – Derek Holt Jun 1 '20 at 21:32
  • $\begingroup$ Can I ask a clarification question? When you write $G=\langle A, B\rangle$, is this a different $G$ from the normaliser? I cannot see any reason why the two subgroups would generate the normaliser. (The proof only uses that $G$ is a subgroup of the normaliser, so this isn't an error, it's just unclear to me what you're meaning.) $\endgroup$ – user1729 Jul 18 at 10:33
  • $\begingroup$ @user1729 I believe that the argument proves that, if $G$ is a minimal counterexample to the claim that no such group exists, then $G = \langle A,B \rangle$ and $A,B \unlhd G$. So I think the answer to your question is that we have proved (under this assumption) that $G = \langle A,B \rangle$ and $A,B \unlhd G$. So yes that means that $G$ is equal to the normalizer. The problem with all this is that this group $G$ doesn't exist anyway, so it is difficult to be definitive about its properties! $\endgroup$ – Derek Holt Jul 18 at 11:32
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As you have correctly pointed out that each cyclic subgroup will have exactly 4 elements of order 10.

claim:- if ${H_k}\cap{H_l}$ contains an element (say) $a_i$ of order 10 then $H_k=H_l$

Since ${a_i}\in{H_k}\cap{H_l}$ and order of $a_i$ is 10 then, $<a_i>=H_k$ and $<a_i>=H_l$ Hence $H_k=H_l$

Did you got why there will be only two cyclic subgroup of order 10?

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