What elements kill $1-\zeta_n$ in the group ring $\mathbb{Z}[G]$? Let $n$ be an integer not congruents to $2$ modulo $4$, also not a prime power.
Let $G$ be the Galois group of $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$.
A fact that I know is $n({\sigma_i}-{\sigma_{-i}})$ and $\sum_{\sigma \in G}{\sigma}$  kill $(1-\zeta_n)$ where $\sigma_i$ is the map sending $\zeta_n$ to $\zeta_{n}^i$.  
I wonder if there are any other types of elements except those.
Help me please.   
 A: Contrary to @user45765, I doubt that the classical proof of Hilbert's thm.90 could give an easy hold on the kernel of the norm in the general case. Let $K/k$ be a finite Galois extension of fields with group $G$. I stick to your notations, which are additive in the group algebra $\mathbf Z[G]$ resp. multiplicative in the group $K^*$. The norm $N$ is the sum of all $\sigma\in G$. Introduce the augmentation map $\alpha: \mathbf Z[G] \to \mathbf Z$ which associates to an element of $\mathbf Z[G]$ the sum of its coefficients; Ker $\alpha:=I_G=I$ is called the augmentation ideal, which is easily shown to be generated by the elements $\sigma-1$ for all $\sigma \in G$. If $G=<s>$ is cyclic, then $I.K^*$ is just the subgroup of $K^*$ consisting of all $\sigma(x)/x$ for all $x\in K^*$, and the usual Hilbert's 90 can be rewritten as Ker $N=I.K^*$. Its generalization to all $G$, which is of a cohomological nature, states that $H^1(G, K^*)=0$, but unfortunately $H^1(G, K^*)\neq Ker N/I.K^*:=H^{-1}(G, K^*)$ (if you don't know about cohomology groups, just take these as notational "black boxes"), the equality between $H^1$ and $H^{-1}$ occurs if $G$ is cyclic, but not always. All one can add in general is that the order $g$ of $G$ kills all the cohomology groups $H^j (G,A)$ for any $G$-module $A$ and any $j\neq0$ (see e.g. Serre, "Local Fields", chap.8, §3 ; for $j=0$, the group $H^0$ needs to be modified), so in particular $gKer N$ is contained in $I.K^*$.
Using class-field theory, one can go a bit further when dealing with non archimedean local fields or with number fields. In the latter case, add the hypothesis that the real places of $k$ do not split in $K$ (this corresponds to your condition that $n\neq 2$ mod $4$). If $G$ is abelian, one can describe explicitly $H^{-1}(G, K^*)$ in terms of $G$ alone. More precisely, let us decompose $G$ as $Z_1 \times Z_2 \times...\times Z_m$, where the groups $Z_j$ are cyclic of orders $n_j$, with $n_j\mid n_{j-1}$ (this is unique up to isomorphism). Writing $A_j= Z_j \times Z_{j+1} \times...\times Z_m$, the main result is an isomorphism of groups $H^{-1}(G, K^*) \cong A_2 \times A_3\times ...\times A_m$, see e.g. thm.1 of the report www.numdam.org/article/SDPP_1975-1976__17_2_A20_0.pdf. This implies that the order of $Z_2$ kills $H^{-1}(G, K^*)$. Another corollary is that $H^1(G, K^*)=H^{-1}(G, K^*)$ iff $G$ is cyclic.
Let us now study the determination (in principle!) of the set $Ann_{\Lambda}(x)=Ann(x)$ of annihilators of a given $x\in Ker N$. This is actually a two-sided ideal of $\Lambda$. For clarity, let us first solve the case of a cyclic extension $K/k$ with group  $G=<\sigma>$. Then any $x\in KerN$ can be written as $x=(\sigma -1)y$, where $y\in K^*$ is defined up to multiplication by a factor in $k^*$. An annihilator $\alpha \in Ann(x)$ is characterized by $\alpha (\sigma-1)y=1$, or $(\sigma-1)(\alpha(y))=1$ by commutativity, or $\alpha(y)\in k^*$. In the case of an abelian extension of local fields or number fields as above, let $Z_j=<\sigma_j>$. For $x\in Ker N$, we saw that   $x^{a_2}$ can be written as a product $\prod (\sigma_j -1)y_j$, with $y_j \in K$, hence $(\prod (\sigma_j -1)\alpha(y_j)^{a_2})=1$ as before, but this sole equation does not suffice to conclude, we need to know some relations between the $y_j$. These should be provided by the explicit description of the groups $A_j$ in the lemma after thm.2.2 of the report cited above, which deals explicitly with the case of a bicyclic extension. Of course things should be simpler in your particular case of a non cyclic cyclotomic field  ./.
