Which of the following statements are necessarily true for a group of order 2022?
(a) Let g be an element of odd order in $G$ and $s_g$ the permutation of $G$ given by $s_g(x)=gx$ for $x\in G$.Then $s_g$ is an even permutation.
(b)The set $H$={$g\in G$ $|$ $o(g)$ is odd} is a normal subgroup in G.
(c)G has a normal subgroup of index 337.
(d) G has only 2 normal subgroups.
Here,$2022=2\times 3\times 337$. For option (c),by third sylow theorem the number of sylow-337-subgroups divides 2022,the only possible value for the number of sylow-337-subgroup is 1.Since this is unique it is normal in G. Then the quotient group $G\over N$ is a group of order 6.But how to show that this subgroup of order 6 is normal in G?
For option (d) the subgroup of index 2 is normal in G and by above sylow-337-subgroup is normal in G.Also there is always two normal subgroups of G which is trivial and improper so option (d) is false.
For option(b) H contains elements of odd order in G.Since the order of element divides order of the group G.The order of each element must be either 3 or 337.(Is this reason correct?).If it is correct then the order of H must be $lcm(337,3)=1011 \leq 2022$.Since index of group of order 1011 is 2.It must be normal subgroup.Also any subgroup of order greater than 1011 cannot be a subgroup by Lagrange theorem.Thus (b) must be true
And for option (a),let $G$={$g_1$,$g_2$,....$g_{2022}$} and let $g=g_{10}$$(say)$.Then $s_{g_{10}}(x)$=$g_{10}x$ for all $x\in G$.Since o($g_{10}$)= odd $\implies$ ${g_{10}}^{2k+1}$=$e$,where $e$ is the identity.Then ${s_{g_{10}}(x)}^2$=$g_{10}g_{10}x$, continuing like this, ${s_{g_{10}}(x)}^{2k+1}$=$ex$=$x$. Thus $o(s_{g_{10}}(x))$= odd.So,it can be written as product of even number of transpositions.Therefore (a) is true.
Anyone could give me a hint for (c)? Or how to show that the quotient group $G \over N$ is normal?
