The structure of the relative SK₁ group of the integral group ring is crucial to compute the K₂ group of the integral group ring. However it is very difficult to determine the exact structure of the relative SK₁ group. We can only find some results about the structure of the relative SK₁ group K₁($\mathbb{Z}$[G]) when G is an elementary Abelian p-group. In this paper,we consider the case ofG=C₄×C₄=<σ,τ|σ⁴=τ⁴=1,στ=τσ>. We use the a result about the exponent of K₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄]) in Ref.^[1] to determine the exact structure of the group K₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄]). We use the result about the structure of K₂(F₂[C₄×C₄]) in Ref.^[2] to study the structure ofK₂($\mathbb{Z}$[C₄×C₄]),and we obtain a lower bound of the 4-rank ofK₂$\mathbb{Z}$[C₄×C₄]) and a lower bound of the 2-rank ofK₂($\mathbb{Z}$[C₄×C₄]).

1 Preliminaries

Let C_n be a cyclic group of order n,p be a prime number,and F_p be a finite field of p elements. Let J(R) denote the Jacobson radical of a ring R. We first introduce a basic definition which is crucial to compute the relative SK₁ group.

Definition 1.1 Let G be a finite Abelian p-group and O_F be the ring of integers in an algebraic number field F. A subset $S\subset \hat{G}$ is called an F-cluster of characters for G if S contains exactly one character for each simple F[G]-module. When F=Q,O_F=begin{document} $mathbb{Z}$ end{document}, we will call S a cluster of characters for G. If S is a cluster of characters for G,We will define S₀=$\left\{ \chi \in S\left| \mathbb{Z}\left[ \chi \right] \right.\ne \mathbb{Z} \right\}$. S₀ will be called an imaginarycluster of characters for G. Note that S₀=S-trivial character} when p is odd and S₀={χ∈S||im(χ)|>2} when p=2.

The following two theorems will be used to determine K₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄]).

Theorem 1.1 (Theorem 1.10 in Ref.^[1]) Let G be a finite Abelian 2-group and S₀ an imaginary cluster of characters for G. Then

$\begin{align} & S{{K}_{1}}\left( \mathbb{Z}\left[ G \right],2\mathbb{Z}\left[ G \right] \right)= \\ & \left[ \underset{\chi \in {{S}_{0}}}{\mathop{\Pi }}\,\operatorname{Im}\chi \right]/{{\psi }_{\chi {{s}_{0}}}}\left( G\otimes \left( 2-\phi \right)\left( J\left( {{{\hat{\mathbb{Z}}}}_{2}}\left[ G \right] \right) \right) \right) \\ \end{align}$

Note The elements $\sum {{\lambda }_{g}}g\in {{{\hat{\mathbb{Z}}}}_{2}}\left[ G \right]$,Φ is defined as follows：

$\left( \sum {{\lambda }_{g}}g \right)=\sum {{\lambda }_{g}}{{g}^{2}}$

∀g,h∈G,ψ_χij is defined as follows:

${{\psi }_{\chi ij}}\left( g\otimes h \right)=\left\{ \begin{align} & \chi ij\left( g \right), if\chi ij\left( h \right)=1 \\ & 1, if\chi ij\left( h \right)\ne 1 \\ \end{align} \right.$

Theorem 1.2 (Proposition 4.7 in Ref.^[1])

exp(SK₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄])=2.

2 Main results

The following are the main results of this paper.

Theorem 2.1 K₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄])=C₂³.

Proof Let G=C₄×C₄=<σ,τ|σ⁴=τ⁴=1,στ=τσ>, and ξ be a primitive 4th root of unity. Define the characters of C₄×C₄ as follows:

χ_ij,0≤i,j≤3,

χ_ij(σ)=ξⁱ,χ_ij(τ)=ξ^j,

then χ_ij(σ^hτ^k)=ξ^hi+jk. Let S₀={χ_1j,0≤j≤3,χ_i1,i=0,2}, then by Proposition 4.7 of Ref.^[1],S₀ is an imaginary cluster of C₄×C₄.

Then by Theorem 1.1,

$\begin{align} & S{{K}_{1}}\left( \mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right],2\mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right)= \\ & \left[ \underset{\chi \in {{S}_{0}}}{\mathop{\Pi }}\,im\chi \right]/{{\psi }_{\chi {{S}_{0}}}}\left( G\otimes \pm 2-\phi \right)\left. \left( J\left( {{{\hat{\mathbb{Z}}}}_{2}}\left[ G \right] \right) \right) \right) \\ \end{align}$

Obviously,im χ₀₁=im χ₁₀=im χ₁₁=im χ₁₂=im χ₁₃=im χ₂₁=C₄. In the following,we will fix the order of the product $\underset{\chi \in {{S}_{0}}}{\mathop{\Pi }}\,$im χ as $\underset{\chi \in {{S}_{0}}}{\mathop{\Pi }}\,$im χ=im χ₀₁×im χ₁₀×im χ₁₁×im χ₁₂×im χ₁₃×im χ₂₁=C₄⁶.

Next we will determine the structure of ${{\psi }_{\chi {{S}_{0}}}}\left( G\otimes \pm 2-\phi \right)\left. \left( J\left( {{{\hat{\mathbb{Z}}}}_{2}}\left[ G \right] \right) \right) \right)$.

For any $x\in J\left( {{{\hat{\mathbb{Z}}}}_{2}}\left[ G \right] \right)=<2,g-1\left| g\in G> \right.$, we have ${{\psi }_{\chi {{S}_{0}}}}\left( \sigma \tau \otimes \left( 2-\phi \right)\left( x \right) \right)={{\psi }_{\chi {{S}_{0}}}}\left( \sigma \otimes \left( 2-\phi \right)\left( x \right) \right)\cdot {{\psi }_{\chi {{S}_{0}}}}\left( \tau \otimes \left( 2-\phi \right)\left( x \right) \right)$, then ${{\psi }_{\chi {{S}_{0}}}}\left( G\otimes \pm 2-\phi \right)\left. \left( J\left( {{{\hat{\mathbb{Z}}}}_{2}}\left[ G \right] \right) \right) \right)$ is generated by the following 34 elements,

$\begin{align} & \left\{ {{\psi }_{\chi {{S}_{0}}}} \right.\left( \sigma \otimes \left( 2-\phi \right)\left( 2 \right) \right),{{\psi }_{\chi {{S}_{0}}}}\left( \tau \otimes \left( 2-\phi \right)\left( 2 \right) \right), \\ & {{\psi }_{\chi {{S}_{0}}}}\left( \sigma \otimes \left( 2-\phi \right)\left( {{\sigma }^{i}}{{\tau }^{j}}-1 \right) \right), \\ & {{\psi }_{\chi {{S}_{0}}}}\left( \tau \otimes \left( 2-\phi \right)\left( {{\sigma }^{i}}{{\tau }^{j}}-1 \right) \right),0\le i,j\le \left. 3 \right\} \\ \end{align}$

We will compute the value of ${{\psi }_{\chi {{S}_{0}}}}\left( \sigma \otimes \left( 2-\phi \right)\left( \sigma \tau -1 \right) \right)$ as an example,and the same method can be used to determine the values of the other 33 elements. For

$\begin{align} & \left( 2-\phi \right)\left( \sigma \tau -1 \right)=2\sigma \tau -2-{{\sigma }^{2}}{{\tau }^{2}}+1 \\ & =-{{\sigma }^{2}}{{\tau }^{2}}+2\sigma \tau -1 \\ \end{align}$

we have

$\begin{align} & {{\psi }_{\chi {{S}_{0}}}}\left( \tau \otimes \left( 2-\phi \right)\left( \sigma \tau -1 \right) \right)= \\ & {{\psi }_{\chi {{S}_{0}}}}\left( \tau \otimes \left( -{{\sigma }^{2}}{{\tau }^{2}}+2\sigma \tau -1 \right) \right)= \\ & \left( {{y}_{ij}} \right)\in \underset{\chi \in {{S}_{0}}}{\mathop{\Pi }}\,im\chi \\ \end{align}$

where

$\begin{align} & {{y}_{ij}}={{\psi }_{\chi ij}}\left( \sigma \otimes \left( -{{\sigma }^{2}}{{\tau }^{2}}+2\sigma \tau -1 \right) \right) \\ & ={{\psi }_{\chi ij}}{{\left( \sigma \otimes {{\sigma }^{2}}{{\tau }^{2}} \right)}^{-1}}{{\psi }_{\chi ij}}{{\left( \sigma \otimes \sigma \tau \right)}^{2}}\times \\ & {{\psi }_{\chi ij}}{{\left( \sigma \otimes 1 \right)}^{-1}} \\ \end{align}$

By the definition of ψ_χij,we have y₀₁=1,y₁₀=ξ³,y₁₁=ξ²,y₁₂=ξ³,y₁₃=1,and y₂₁=ξ². Hence

ψ_χS0(σ⊗(2-φ)(στ-1))=(1,ξ³,ξ²,ξ³,1,ξ²).

Using the same method,we can get the following generators:

ψ_χS0(σ⊗(2-φ)(1-1)

=ψ_χS0(τ⊗(1-1))=(1,1,1,1,1,1),

{ψ_χS0(σ⊗(2-φ)(2))=(1,ξ²,ξ²,ξ²,ξ²,1),

ψ_χS0(τ⊗(2-φ)(2))=(ξ²,1,ξ²,1,ξ²,ξ²),

ψ_χS0(σ⊗(2-φ)(τ-1))=(1,1,ξ³,ξ²,ξ³,ξ²),

ψ_χS0(σ⊗(2-φ)(τ²-1))=(1,1,ξ²,1,ξ²,1),

ψ_χS0(σ⊗(2-φ)(τ³-1))=(1,1,ξ³,ξ²,ξ³,ξ²),

ψ_χS0(σ⊗(2-φ)(σ-1))=(1,ξ³,ξ³,ξ³,ξ³,1),

ψ_χS0(σ⊗(2-φ)(στ-1))=(1,ξ³,ξ²,ξ³,1,ξ²),

ψ_χS0(σ⊗(2-φ)(στ²-1))=(1,ξ³,ξ³,ξ³,ξ³,1),

ψ_χS0(σ⊗(2-φ)(στ³-1))=(1,ξ³,1,ξ³,ξ²,ξ²),

ψ_χS0(σ⊗(2-φ)(σ²-1))=(1,ξ²,ξ²,ξ²,ξ²,1),

ψ_χS0(σ⊗(2-φ)(σ²τ-1))=(1,ξ²,ξ³,1,ξ³,ξ²),

ψ_χS0(σ⊗(2-φ)(σ²τ²-1))=(1,ξ²,1,ξ²,1,1),

ψ_χS0(σ⊗(2-φ)(σ²τ³-1))=(1,ξ²,ξ³,1,ξ³,ξ²),

ψ_χS0(σ⊗(2-φ)(σ³-1))=(1,ξ³,ξ³,ξ³,ξ³,1),

ψ_χS0(σ⊗(2-φ)(σ³τ-1))=(1,ξ³ξ³,1,ξ³,ξ²,ξ²),

ψ_χS0(σ⊗(2-φ)(σ³τ²-1))=(1,ξ³ξ³,ξ³,ξ³,ξ³,1),

ψ_χS0(σ⊗(2-φ)(σ³τ³-1))=(1,ξ³,ξ²,ξ³,1,ξ²),

ψ_χS0(τ⊗(2-φ)(τ-1))=(ξ³,1,ξ³,1,ξ,ξ³),

ψ_χS0(τ⊗(2-φ)(τ²-1))=(ξ²,1,ξ²,1,ξ²,ξ²),

ψ_χS0(τ⊗(2-φ)(τ³-1))=(ξ³,1,ξ³,1,ξ,ξ³),

ψ_χS0(τ⊗(2-φ)(σ-1))=(1,1,ξ³,ξ²,ξ,ξ²),

ψ_χS0(τ⊗(2-φ)(στ-1))=(ξ³,1,ξ²,ξ²,1,ξ³),

ψ_χS0(τ⊗(2-φ)(στ²-1))=(ξ²,1,ξ³,ξ²,ξ,1),

ψ_χS0(τ⊗(2-φ)(στ³-1))=(ξ³,1,1,ξ²,ξ²,ξ³),

ψ_χS0(τ⊗(2-φ)(σ²-1))=(1,1,ξ²,1,ξ²,1),

ψ_χS0(τ⊗(2-φ)(σ²τ-1))=(ξ³,1,ξ³,1,ξ,ξ³),

ψ_χS0(τ⊗(2-φ)(σ²τ²-1))=(ξ²,1,1,1,1,ξ²),

ψ_χS0(τ⊗(2-φ)(σ²τ³-1))=(ξ³,1,ξ³,1,ξ,ξ³),

ψ_χS0(τ⊗(2-φ)(σ³-1))=(1,1,ξ³,ξ²,ξ,ξ²),

ψ_χS0(τ⊗(2-φ)(σ³τ-1))=(ξ³,1,1,ξ²,ξ²,ξ³),

ψ_χS0(τ⊗(2-φ)(σ³τ²-1))=(ξ²,1,ξ³,ξ²,ξ,1),

ψ_χS0(τ⊗(2-φ)(σ³τ³-1))=(ξ³,1,ξ²,ξ²,1,ξ³)}.

Because some of these generators are the same,we use a_i to denote the different generators：

{a₁=(1,ξ²,ξ²,ξ²,ξ²,1);

a₂=(ξ²,1,ξ²,1,ξ²,ξ²);

a₃=(1,1,ξ³,ξ²,ξ³,ξ²);

a₄=(1,1,ξ²,1,ξ²,1);

a₅=(1,ξ³,ξ³,ξ³,ξ³,1);

a₆=(1,ξ³,ξ²,ξ³,1,ξ²);

a₇=(1,ξ³,1,ξ³,ξ²,ξ²);

a₈=(1,ξ²,ξ³,1,ξ³,ξ²);

a₉=(1,ξ²,1,ξ²,1,1);

a₁₀=(ξ³,1,ξ³,1,ξ,ξ³);

a₁₁=(1,1,ξ³,ξ²,ξ,ξ²);

a₁₂=(ξ³,1,ξ²,ξ²,1,ξ³);

a₁₃=(ξ²,1,ξ³,ξ²,ξ,1);

a₁₄=(ξ³,1,1,ξ²,ξ²,ξ³);

a₁₅=(ξ²,1,1,1,1,ξ²);

a₁₆=(ξ³,1,ξ²,ξ²,1,ξ³).}

Next we will determine the structure of the group generated by a_i which is ψ_χS0(G⊗(2-φ)(J(${\hat{\mathbb{Z}}}$₂[G]))).

For 1≤i≤6,let b_i be the vector of dimension 6 in which the ith component is ξ² and the other components are all equal to 1. Let b₇=(ξ,1,1,1,1,ξ),b₈=(1,ξ,1,ξ,1,1),b₉=(1,1,ξ,1,ξ,1). We show that {b_i,1≤i≤9} is a generating set of the group generated by {a_i,1≤i≤16}.

On one hand,{b_i,1≤i≤9} can be generated by {a_i,1≤i≤16}:

a₅·a₆·a₁₁=(1,ξ²,1,1,1,1)=b₂,

a₉·b₂=(1,1,1,ξ²,1,1)=b₄,

a₅·a₇·a₈·b₄=(1,1,ξ²,1,1,1)=b₃,

a₄·b₃=(1,1,1,1,ξ²,1)=b₅,

a₃·a₄·a₁₀·a₁₂·a₁₄·a₁₆=(1,1,1,1,1,ξ²)=b₆,

a₁₅·b₆=(ξ²,1,1,1,1,1)=b₁,

a₁₂·b₁·b₃·b₄·b₆=(ξ,1,1,1,1,ξ)=b₇,

a₇·b₂·b₄·b₅·b₆=(1,ξ,1,ξ,1,1)=b₈,

a₈·b₂·b₃·b₅·b₆=(1,1,ξ,1,ξ,1)=b₉.

On the other hand,{a_i,1≤i≤16} can be generated by {b_i,1≤i≤9}:

a₁=b₂·b₃·b₄·b₅,a₂=b₁·b₃·b₅·b₆,

a₃=b₃·b₄·b₅·b₆·b₉,a₄=b₃·b₅,

a₅=b₂·b₃·b₄·b₅·b₈·b₉,a₆

=b₂·b₃·b₄·b₆·b₈,

a₇=b₂·b₄·b₅·b₆·b₈,a₈=b₂·b₃·b₅·b₆·b₉,

a₉=b₂·b₄,a₁₀=b₁·b₃·b₆·b₇·b₉,

a₁₁=b₃·b₄·b₆·b₉,a₁₂=b₁·b₃·b₄·b₆·b₇,

a₁₃=b₁·b₃·b₄·b₉,a₁₄=b₁·b₄·b₅·b₆·b₇,

a₁₅=b₁·b₆,a₁₆=b₁·b₃·b₄·b₆·b₇.

Hence {b_i,1≤i≤9} is an generating set of ψ_χS0(G⊗(2-φ)(J(${\hat{\mathbb{Z}}}$₂[G]))).

It is easy to know that these 6 elements {b_i,1≤i≤6} generate an elementary Abelian 2-group of rank 6. We denote this group by H. Then the 8 cosets,

{H,b₇H,b₈H,b₉H,b₇b₈H,b₇b₉H,b₈b₉H,b₇b₈b₉H}，

are disjoint with each other and their union is ψ_χS0(G⊗(2-φ)(J(${\hat{\mathbb{Z}}}$₂[G]))). Hence

|ψ_χS0(G⊗(2-φ)(J(${\hat{\mathbb{Z}}}$₂[G])))|=2⁶·8=2⁹,

and

$\left| S{{K}_{1}}\left( \mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right],2\mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right) \right|\frac{{{4}^{6}}}{{{2}^{9}}}={{2}^{3}}$

By Theorem 1.2,

exp(SK₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄])=2.

So

K₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄])=C₂³.

□

Corollary 2.1 K₁($\mathbb{Z}$[C₄×C₄×C₂])=C₂⁴.

Proof By Theorem 1.11 in Ref.^[1],we have K₁($\mathbb{Z}$[C₄×C₄×C₂])⊕SK₁($\mathbb{Z}$[C₄×C₄)SK₁($\mathbb{Z}$［C₄×C₄],2$\mathbb{Z}$[C₄×C₄]). By Theorem 5.5 in Ref.^[1],K₁($\mathbb{Z}$[C₄×C₄])=C₂ . Then by Theorem 2.1,we have

K₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄])=C₂³.

So K₁($\mathbb{Z}$[C₄×C₄×C₂])=C₂⁴. □

Theorem 2.2 The 4-rank ofK₂($\mathbb{Z}$[C₄×C₄]) is at least 1 and its 2-rank is at least 10.

Proof By the long exact sequence of K-theory,we have

$\begin{align} & {{K}_{2}}\left( \mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right)\xrightarrow{{{f}_{1}}}{{K}_{2}}\left( {{\mathbb{Z}}_{2}}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right)\xrightarrow{{{f}_{2}}} \\ & S{{K}_{1}}\left( \mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right),2\mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right]\xrightarrow{{{f}_{3}}} \\ & S{{K}_{1}}\left( \mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right)\xrightarrow{{{f}_{4}}}S{{K}_{1}}\left( {{F}_{2}}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right) \\ \end{align}$

By Theorem 1.2 in Ref.^[2], K₂(F₂[C₄×C₄])=C₂⁹⊕C₄³. By Theorem 2.1,

K₁($\mathbb{Z}$[C₄×C₄],2$\mathbb{Z}$[C₄×C₄])=C₂³. By Theorem 5.5 in Ref.^[1],we have K₁($\mathbb{Z}$[C₄×C₄])=C₂. Then the exact sequence becomes

$\begin{align} & {{K}_{2}}\left( \mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right)\xrightarrow{{{f}_{1}}}C_{2}^{9}\oplus C_{4}^{3}\xrightarrow{{{f}_{2}}} \\ & C_{2}^{3}\xrightarrow{{{f}_{3}}}{{C}_{2}}\xrightarrow{{{f}_{4}}}1 \\ \end{align}$

By the exactness, im(f₂)=ker(f₃)=C₂². Then we get the following exact sequence,

$\begin{align} & {{K}_{2}}\left( \mathbb{Z}\left[ {{C}_{4}}\times {{C}_{4}} \right] \right)\xrightarrow{{{f}_{1}}}C_{2}^{9}\oplus C_{4}^{3}\xrightarrow{{{f}_{2}}} \\ & C_{2}^{2}\to 1 \\ \end{align}$

By the exactness, im(f₁)=ker(f₂) can only be one of the following three cases,

{C₂¹¹⊕C₄;C₂⁹⊕C₄²;C₂⁷⊕C₄³}.

In any of these cases, im(f₁) contains one cyclic subgroup of order 4 as its direct summand. Then the 4-rank ofK₂($\mathbb{Z}$[C₄×C₄]) is at least 1 and the 2-rank is at least 10. □