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Please show that $A_{5}$, a group of order $60$, has no subgroup of order $15$.

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    I know A_{5} is a simple group2012-04-23
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    @Ali That's all you know about $A_5$? Do you not know what the elements are? What the elements of order 3 and 5 are?2012-04-23

3 Answers 3

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Show that every group of order $15$ is cyclic. The result follows since there is no element of order $15$ in $A_5$.

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Hint

  • $A_5$ is simple.

  • What is the index of such a group? Let $A_5$, a simple group act on left cosets of this proper subgroup? What can you say about the kernel of the homomorphism that comes with this action?

  • So, now apply first isomorphism theorem; Lagrange's theorem to conclude a result known due to Poincare...

  • So, what do you conclude?


Perhaps, a more adhoc solution that applies exclusively here, but nonetheless, an important fact would be to prove the following:

  • $A_5$ has no element of order $15$. (Perhaps, you should try to list all those orders that occur in $A_5$.)

  • A group of order $15$ is cyclic. (Perhaps, I suggest you classify groups of order $pq$ for primes $p$ and $q$. This is a fun exercise and I suggest you'll do this. You'll get comfortable thinking about group actions and Sylow's theorem. )

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    Please explain more2012-04-23
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    there is a homomorphism: $G$ to sym(4). Hence $G$ has a non trival normal subgroup that is contradiction2012-04-23
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    @Ali You're uttering the right words in wrong order. Perhaps, try a bit more harder. You're closer to the answer. Think a bit more before I'll tell you anything, Regards,2012-04-23
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    I don't think the second solution is ad hoc, the exact same argument works when $A_5$ is replaced by $S_n$ or $A_n$ for $n \leq 7$.2012-04-23
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    If $H\subseteq G$ and $[G:H]=n$, then there is a homorphism of $G$ to $S_{n}$ such that $ker\subseteq H$.2012-04-23
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    Right, so, what can the kernel be?2012-04-23
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    Do we can to say that kernel is non trivial?2012-04-23
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    Can it be non-trivial, remember simple groups...So, no non-trivial $\underline{{}{}{}}$ subgroups...2012-04-23
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    ok. hence kernel is trivial. Then order $G$(60) divides order $S_{4}$ (24), that is a contradiction.2012-04-23
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    @Ali You're right about the contradiction. Now generalize this result and make it a part of your tool set. I am only an undergraduate student. Would you mind deleting your previous comment? Regards,2012-04-23
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Assume $H \leq A_5$ with $|H| = 15$ and let $X:=\{gH \mid g \in G\}$. Then $\# X = 4$. $G$ acts op $X$ by left multiplication i.e. $g'(gH) = (g'g)H$. Let $\alpha \in A_5$ be a 5-cycle. Then $\langle \alpha\rangle$ does act on $X$,too. But the length of an orbit divides the group-order which is 5. But $\# X = 4 < 5$ so each orbit contains only one element. That means $\alpha H = H$ for all $\alpha$. So $\alpha \in H$. There are 24 of those $\alpha$. Contradition because $H$ cannot contain more than 15 elements.