Please show that $A_{5}$, a group of order $60$, has no subgroup of order $15$.
Why $A_{5}$ has no subgroup of order $15$?
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0I know A_{5} is a simple group – 2012-04-23
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0@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
Show that every group of order $15$ is cyclic. The result follows since there is no element of order $15$ in $A_5$.
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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0Please explain more – 2012-04-23
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0there is a homomorphism: $G$ to sym(4). Hence $G$ has a non trival normal subgroup that is contradiction – 2012-04-23
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0@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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0I 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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0If $H\subseteq G$ 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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0Right, so, what can the kernel be? – 2012-04-23
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0Do we can to say that kernel is non trivial? – 2012-04-23
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0Can it be non-trivial, remember simple groups...So, no non-trivial $\underline{{}{}{}}$ subgroups... – 2012-04-23
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0ok. hence kernel is trivial. Then order $G$(60) divides order $S_{4}$ (24), that is a contradiction. – 2012-04-23
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0@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
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.