Prove that every group of order 15 is abelian?

Question

I had seen this proof at many places, but everywhere sylows theorem is used.

So is their any way to solve it without using sylows theorem?

Answer 1

Let $G$ be a group of order $15$. There exist an element $a$ of order $5$ and an element $b$ of order $3$. Now we consider $b^{-1}ab$, and we have $(b^{-1}ab)^5 = 1$. So $o(b^{-1}ab)\mid 5$, which implies $o(b^{-1}ab) = 1$ or $5$. It cannot be $1$, otherwise $ab = b$ and so $a = 1$, a contradiction. Thus $o(b^{-1}ab) = 5$.

We claim that $b^{-1}ab = a^m$ for some $m$. Otherwise $\langle a\rangle$, $\langle b^{-1}ab\rangle$, $\langle bab^{-1} = b^{-2}ab^2\rangle$ are three subgroups of $G$ of order $5$, each two of them intersect trivially. So we have $$G = \{1,a,\dots,a^4,b^{-1}ab,\dots,(b^{-1}ab)^4,b^{-2}ab^2,\dots,(b^{-2}ab^2)^4,b,b^2\}.$$ However, you may check that $ba$ is not in this set.

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To check $ba\ne a^k,ba^kb^{-1},b^k$ is easy, we now check $ba\ne b^{-1}a^kb$.

For $k = 1$, we have $aba^{-1}=b^2=b^{-1}$, then $a^2b=ba^2$. Hence $b,a^2$ commutative and thus $b,a = (a^2)^3$ commutative.

For $k = 2$, $bababa = b^{-1}ab$ and so $b^{-1}ab = (ba^2ba)^2$, we have an element of order $10$, a contradiction.

For $k=3$, $baba = b^{-1}ab$ and so $(b^2ab^2)^2 = a$.

For $k = 4$, then $a^{-1}b^{-1} = (ba)^{-1} = b^{-1}ab$ and so $(ab^{-1})^2 = 1$.

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Note that $o(b) = 3$, and so $m^3\equiv 1\pmod 5$. The only solution of that equation is $m\equiv 1\pmod 5$. Therefore, $b^{-1}ab = a$. Now $ab$ is of order $15$, $G = \langle ab\rangle$ is cyclic.

Answer 2

Sure. Instead of Sylow's theorems we can get normality of a subgroup by the following: if $p$ is the smallest prime dividing $|G|$ then any subgroup of index $p$ is normal in $G$ (you can consider this an exercise if you want, but a proof can be found here). For us, $|G|=15$ so $p=3$.

Now, there exists an element $g\in G$ of order $5$ (why?). This generates a subgroup $N$, and $N$ is normal by the above. Consider $G/N$, which is cyclic of order $3$. Pick some element $h\not\in N$. The cyclicity of $GN$ means that if $h\not\in N$ then $h^3\in N$.

Suppose $h^3\neq1$. Then $h$ has order $15$ and $G$ is cyclic (and hence abelian).

Suppose $h^3=1$. Then we have to use semidirect products: $G$ splits as a semidirect product $N\rtimes_{\phi}G/N$, where $\phi: G/N\rightarrow\operatorname{Aut}(N)$. Recalling all the isomorphisms, we have: $\mathbb{Z}_5\rtimes_{\phi}\mathbb{Z}_3$, where $\phi: \mathbb{Z}_3\rightarrow\operatorname{Aut}(\mathbb{Z}_5)$. Now, $\operatorname{Aut}(\mathbb{Z}_5)\cong\mathbb{Z}_4$ (why?) so $\phi$ is the trivial map. Hence, $G\cong \mathbb{Z}_5\times\mathbb{Z}_5\cong\mathbb{Z}_{15}$, and so $G$ is cyclic (and hence abelian).

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I think Sylow's theorems are more low-level than semidirect products. It would be nice to see a proof which didn't use semidirect products :-)

Answer 3

Let $X_k$ be the set of elements of $G$ of order $k$. We know that $X_3,X_5$ are non-empty. Assume $X_{15}=\emptyset$. As distinct subgroups of prime order have trivial intersection, we conclude that $|X_5|$ is a multiple of $4$ and $|X_3|$ is a multiple of $2$. Thus $|X_3\cup X_5|=14$ is only possible if $(|X_3|,|X_5|)$ is one of $(2,12)$, $(6,8)$, $(10,4)$. In the first two cases, $|X_3|$ is not a multiple of $5$, hence the action by conjugation of any element $b\in X_5$ has a fixedpoint $a\in X_3$. In the last case, $|X_5|$ is not a multiple of $3$, hence the action by conjugation of any $a\in X_3$ has a fixedpoint $b\in X_5$. At any rate we have $ab=ba$, hence $(ab)^3=b^3\ne1$, $(ab)^5=a\ne 1$, contradiction.

We conclude that $X_{15}\ne\emptyset$ and hence $G$ is cyclic, in particular abelian.