Consider a group $G$ of order $10$.
Then $G$ can be abelian as well non-abelian.
What is the number of non-trivial elements of $G$ of order $2$?
Answer:
If $G$ is abelian, $G$ can be cyclic as well non-cyclic.
If $G$ is cyclic ,then it is isomorphic to $\mathbb{Z}_{10}$,which has only $1$ element of order $2$ i.e., $\bar 5$ .
If $G$ is non-cyclic,then also it has $1$ element of order $2$.
But if $G$ be non-abelian,then the situation becomes different.
By Sylow principle, $G$ has $5$ Sylow $2$-subgroups and hence there are $5$ elements of order $2$.
My question is-
Edit: How to turn the non-abelian case when the group will abelian using Sylow theorem??
In non-abelian case we get $5$ sylow $2$-subgroup.
How this turn into $1$ sylow $2$-subgroup is $G$ becomes abelian ??
Best Answer
I am not sure what exactly you want from your unclear question, but I hope this will ease your confusion. I shall argue that the number of Sylow $2$-subgroups in this case is what differentiates abelian groups of order $10$ from nonabelian groups of order $10$.
Let $G$ be a group of order $10$. Suppose that $n_p$ is the number of $p$-Sylow subgroups of $G$, where $p$ is a prime natural number. In what follows, for each positive integer $k$, $C_k$ denotes the cyclic group of order $k$.
By Sylow Theorems, $n_5\equiv 1\pmod{5}$ and $n_5\mid 2$. Therefore, $n_5=1$ is the only possibility. Thus, there is a unique Sylow $5$-subgroup of $G$. This subgroup must be normal, and we shall call it $N$. Since $|N|=5$, we have $N\cong C_5$.
Using Sylow Theorems again, $n_2\equiv 1\pmod{2}$ and $n_2\mid 5$. This leaves two possibilities: $n_2=1$ and $n_2=5$.
If $n_2=1$, then there exists a unique Sylow $2$-subgroup of $G$. Again, this subgroup must be normal, and we shall call it $H$. As $|H|=2$, we have $H\cong C_2$. Since $N\cap H$ is a subgroup of $N$ and $H$, we have $$|N\cap H|\,\Big\vert\,|N|=5\text{ and }|N\cap H|\,\Big\vert\,|H|=2\,.$$ This shows that $$|N\cap H|\,\Big\vert\,\gcd(5,2)=1\,.$$ That is, $|N\cap H|=1$, making $N\cap H=\{1\}$, the trivial subgroup of $G$. This shows that $N$ commutes with $H$. (If $x\in N$ and $y\in H$, then $$xyx^{-1}y^{-1}=(xyx^{-1})y^{-1}\in H\text{ and }xyx^{-1}y^{-1}=x(yx^{-1}y^{-1})\in N\,,$$ whence $xyx^{-1}y^{-1}\in N\cap H=\{1\}$). Therefore, the subgroup $\tilde{G}$ of $G$ generated by $N$ and $H$ is isomorphic to the direct product $N\times H$. Because $N\cong C_5$ and $H\cong C_2$, we get $$\tilde{G}\cong N\times H\cong C_5\times C_2\cong C_{10}$$ is an abelian group of order $10$. As $G$ has order $10$, it follows that $G$ is equal to $\tilde{G}$, making $G$ a cyclic group of order $10$.
Let now assume that $n_2=5$. Suppose that the five Sylow $2$-subgroups of $G$ are $\{1,a\}$, $\{1,b\}$, $\{1,c\}$, $\{1,d\}$, and $\{1,e\}$. Write $N=\{1,x,x^2,x^3,x^4\}$ for some $x\in N$. By the same argument as the previous paragraph, we can see that $a,b,c,d,e\notin N$. Because $|G|=10$, we get $$G=\{1,x,x^2,x^3,x^4,a,b,c,d,e\}\,.$$ We shall now prove that, in this case, $G$ is nonabelian. If $G$ were abelian, then it follows that $\{1,a\}$ is a normal subgroup of $G$ (every subgroup of an abelian group is normal). However, this means $\{1,a\}$ is the only conjugate of $\{1,a\}$. However, other Sylow $2$-subgroups of $G$ are conjugates to $\{1,a\}$, and this yields a contradiction. Therefore, $G$ is nonabelian.
In conclusion, whether $G$ is abelian is dictated by whether $n_2=1$ or $n_2=5$. We can in fact infer from the previous paragraph that there is a unique nonabelian group $G$ of order $10$ up to isomorphism. This group is the dihedral group $D_5$ of order $10$.
To see the last assertion, observe that $axa^{-1}=x^k$ for some $k\in\{0,1,2,3,4\}$. Therefore, $$x=a^2xa^{-2}=a(axa^{-1})a^{-1}=ax^ka^{-1}=(axa^{-1})^k=(x^k)^k=x^{k^2}\,.$$ Thus, $x^{k^2-1}=1$, whence $5\mid k^2-1$. This shows that $k=1$ or $k=4$. If $k=1$, then we can see that $G$ is abelian, which is not what we want. If $k=4$, then $G$ is generated by $x$ and $a$ with $x^5=1$, $a^2=1$, and $axa^{-1}=x^4=x^{-1}$. This is precisely the definition of $D_5$.