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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¡ÁC₄=<¦Ò,¦Ó|¦Ò⁴=¦Ó⁴=1,¦Ò¦Ó=¦Ó¦Ò>. We use the a result about the exponent of K₁($\mathbb{Z}$[C₄¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄]) in Ref.^[1] to determine the exact structure of the group K₁($\mathbb{Z}$[C₄¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄]). We use the result about the structure of K₂(F₂[C₄¡ÁC₄]) in Ref.^[2] to study the structure ofK₂($\mathbb{Z}$[C₄¡ÁC₄]),and we obtain a lower bound of the 4-rank ofK₂$\mathbb{Z}$[C₄¡ÁC₄]) and a lower bound of the 2-rank ofK₂($\mathbb{Z}$[C₄¡ÁC₄]).
1 PreliminariesLet 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={¦Ö¡ÊS||im(¦Ö)|>2} when p=2.
The following two theorems will be used to determine K₁($\mathbb{Z}$[C₄¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄]).
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 $\sum {{\lambda }_{g}}g\in {{{\hat{\mathbb{Z}}}}_{2}}\left[ G \right]$,¦µ is defined as follows£º

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

?g,h¡ÊG,¦×_¦Öij 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¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄])=2.
2 Main resultsThe following are the main results of this paper.
Theorem 2.1 K₁($\mathbb{Z}$[C₄¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄])=C₂³.
Proof Let G=C₄¡ÁC₄=<¦Ò,¦Ó|¦Ò⁴=¦Ó⁴=1,¦Ò¦Ó=¦Ó¦Ò>, and ¦Î be a primitive 4th root of unity. Define the characters of C₄¡ÁC₄ as follows:
¦Ö_ij,0¡Üi,j¡Ü3,
¦Ö_ij(¦Ò)=¦Îⁱ,¦Ö_ij(¦Ó)=¦Î^j,
then ¦Ö_ij(¦Ò^h¦Ó^k)=¦Î^hi+jk. Let S₀={¦Ö_1j,0¡Üj¡Ü3,¦Ö_i1,i=0,2}, then by Proposition 4.7 of Ref.^[1],S₀ is an imaginary cluster of C₄¡ÁC₄.
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 ¦Ö₀₁=im ¦Ö₁₀=im ¦Ö₁₁=im ¦Ö₁₂=im ¦Ö₁₃=im ¦Ö₂₁=C₄. In the following,we will fix the order of the product $\underset{\chi \in {{S}_{0}}}{\mathop{\Pi }}\,$im ¦Ö as $\underset{\chi \in {{S}_{0}}}{\mathop{\Pi }}\,$im ¦Ö=im ¦Ö₀₁¡Áim ¦Ö₁₀¡Áim ¦Ö₁₁¡Áim ¦Ö₁₂¡Áim ¦Ö₁₃¡Áim ¦Ö₂₁=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 ¦×_¦Öij,we have y₀₁=1,y₁₀=¦Î³,y₁₁=¦Î²,y₁₂=¦Î³,y₁₃=1,and y₂₁=¦Î². Hence
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò¦Ó-1))=(1,¦Î³,¦Î²,¦Î³,1,¦Î²).
Using the same method,we can get the following generators:
¦×_¦ÖS0(¦Ò?(2-¦Õ)(1-1)
=¦×_¦ÖS0(¦Ó?(1-1))=(1,1,1,1,1,1),
{¦×_¦ÖS0(¦Ò?(2-¦Õ)(2))=(1,¦Î²,¦Î²,¦Î²,¦Î²,1),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(2))=(¦Î²,1,¦Î²,1,¦Î²,¦Î²),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ó-1))=(1,1,¦Î³,¦Î²,¦Î³,¦Î²),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ó²-1))=(1,1,¦Î²,1,¦Î²,1),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ó³-1))=(1,1,¦Î³,¦Î²,¦Î³,¦Î²),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò-1))=(1,¦Î³,¦Î³,¦Î³,¦Î³,1),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò¦Ó-1))=(1,¦Î³,¦Î²,¦Î³,1,¦Î²),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò¦Ó²-1))=(1,¦Î³,¦Î³,¦Î³,¦Î³,1),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò¦Ó³-1))=(1,¦Î³,1,¦Î³,¦Î²,¦Î²),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò²-1))=(1,¦Î²,¦Î²,¦Î²,¦Î²,1),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò²¦Ó-1))=(1,¦Î²,¦Î³,1,¦Î³,¦Î²),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò²¦Ó²-1))=(1,¦Î²,1,¦Î²,1,1),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò²¦Ó³-1))=(1,¦Î²,¦Î³,1,¦Î³,¦Î²),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò³-1))=(1,¦Î³,¦Î³,¦Î³,¦Î³,1),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò³¦Ó-1))=(1,¦Î³¦Î³,1,¦Î³,¦Î²,¦Î²),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò³¦Ó²-1))=(1,¦Î³¦Î³,¦Î³,¦Î³,¦Î³,1),
¦×_¦ÖS0(¦Ò?(2-¦Õ)(¦Ò³¦Ó³-1))=(1,¦Î³,¦Î²,¦Î³,1,¦Î²),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ó-1))=(¦Î³,1,¦Î³,1,¦Î,¦Î³),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ó²-1))=(¦Î²,1,¦Î²,1,¦Î²,¦Î²),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ó³-1))=(¦Î³,1,¦Î³,1,¦Î,¦Î³),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò-1))=(1,1,¦Î³,¦Î²,¦Î,¦Î²),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò¦Ó-1))=(¦Î³,1,¦Î²,¦Î²,1,¦Î³),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò¦Ó²-1))=(¦Î²,1,¦Î³,¦Î²,¦Î,1),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò¦Ó³-1))=(¦Î³,1,1,¦Î²,¦Î²,¦Î³),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò²-1))=(1,1,¦Î²,1,¦Î²,1),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò²¦Ó-1))=(¦Î³,1,¦Î³,1,¦Î,¦Î³),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò²¦Ó²-1))=(¦Î²,1,1,1,1,¦Î²),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò²¦Ó³-1))=(¦Î³,1,¦Î³,1,¦Î,¦Î³),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò³-1))=(1,1,¦Î³,¦Î²,¦Î,¦Î²),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò³¦Ó-1))=(¦Î³,1,1,¦Î²,¦Î²,¦Î³),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò³¦Ó²-1))=(¦Î²,1,¦Î³,¦Î²,¦Î,1),
¦×_¦ÖS0(¦Ó?(2-¦Õ)(¦Ò³¦Ó³-1))=(¦Î³,1,¦Î²,¦Î²,1,¦Î³)}.
Because some of these generators are the same,we use a_i to denote the different generators£º
{a₁=(1,¦Î²,¦Î²,¦Î²,¦Î²,1);
a₂=(¦Î²,1,¦Î²,1,¦Î²,¦Î²);
a₃=(1,1,¦Î³,¦Î²,¦Î³,¦Î²);
a₄=(1,1,¦Î²,1,¦Î²,1);
a₅=(1,¦Î³,¦Î³,¦Î³,¦Î³,1);
a₆=(1,¦Î³,¦Î²,¦Î³,1,¦Î²);
a₇=(1,¦Î³,1,¦Î³,¦Î²,¦Î²);
a₈=(1,¦Î²,¦Î³,1,¦Î³,¦Î²);
a₉=(1,¦Î²,1,¦Î²,1,1);
a₁₀=(¦Î³,1,¦Î³,1,¦Î,¦Î³);
a₁₁=(1,1,¦Î³,¦Î²,¦Î,¦Î²);
a₁₂=(¦Î³,1,¦Î²,¦Î²,1,¦Î³);
a₁₃=(¦Î²,1,¦Î³,¦Î²,¦Î,1);
a₁₄=(¦Î³,1,1,¦Î²,¦Î²,¦Î³);
a₁₅=(¦Î²,1,1,1,1,¦Î²);
a₁₆=(¦Î³,1,¦Î²,¦Î²,1,¦Î³).}
Next we will determine the structure of the group generated by a_i which is ¦×_¦ÖS0(G?(2-¦Õ)(J(${\hat{\mathbb{Z}}}$₂[G]))).
For 1¡Üi¡Ü6,let b_i be the vector of dimension 6 in which the ith component is ¦Î² and the other components are all equal to 1. Let b₇=(¦Î,1,1,1,1,¦Î),b₈=(1,¦Î,1,¦Î,1,1),b₉=(1,1,¦Î,1,¦Î,1). We show that {b_i,1¡Üi¡Ü9} is a generating set of the group generated by {a_i,1¡Üi¡Ü16}.
On one hand,{b_i,1¡Üi¡Ü9} can be generated by {a_i,1¡Üi¡Ü16}:
a₅¡¤a₆¡¤a₁₁=(1,¦Î²,1,1,1,1)=b₂,
a₉¡¤b₂=(1,1,1,¦Î²,1,1)=b₄,
a₅¡¤a₇¡¤a₈¡¤b₄=(1,1,¦Î²,1,1,1)=b₃,
a₄¡¤b₃=(1,1,1,1,¦Î²,1)=b₅,
a₃¡¤a₄¡¤a₁₀¡¤a₁₂¡¤a₁₄¡¤a₁₆=(1,1,1,1,1,¦Î²)=b₆,
a₁₅¡¤b₆=(¦Î²,1,1,1,1,1)=b₁,
a₁₂¡¤b₁¡¤b₃¡¤b₄¡¤b₆=(¦Î,1,1,1,1,¦Î)=b₇,
a₇¡¤b₂¡¤b₄¡¤b₅¡¤b₆=(1,¦Î,1,¦Î,1,1)=b₈,
a₈¡¤b₂¡¤b₃¡¤b₅¡¤b₆=(1,1,¦Î,1,¦Î,1)=b₉.
On the other hand,{a_i,1¡Üi¡Ü16} can be generated by {b_i,1¡Üi¡Ü9}:
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¡Üi¡Ü9} is an generating set of ¦×_¦ÖS0(G?(2-¦Õ)(J(${\hat{\mathbb{Z}}}$₂[G]))).
It is easy to know that these 6 elements {b_i,1¡Üi¡Ü6} 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 ¦×_¦ÖS0(G?(2-¦Õ)(J(${\hat{\mathbb{Z}}}$₂[G]))). Hence
|¦×_¦ÖS0(G?(2-¦Õ)(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¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄])=2.
So
K₁($\mathbb{Z}$[C₄¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄])=C₂³.
¡õ
Corollary 2.1 K₁($\mathbb{Z}$[C₄¡ÁC₄¡ÁC₂])=C₂⁴.
Proof By Theorem 1.11 in Ref.^[1],we have K₁($\mathbb{Z}$[C₄¡ÁC₄¡ÁC₂])¨’SK₁($\mathbb{Z}$[C₄¡ÁC₄)«ÝSK₁($\mathbb{Z}$£ÛC₄¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄]). By Theorem 5.5 in Ref.^[1],K₁($\mathbb{Z}$[C₄¡ÁC₄])=C₂ . Then by Theorem 2.1,we have
K₁($\mathbb{Z}$[C₄¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄])=C₂³.
So K₁($\mathbb{Z}$[C₄¡ÁC₄¡ÁC₂])=C₂⁴. ¡õ
Theorem 2.2 The 4-rank ofK₂($\mathbb{Z}$[C₄¡ÁC₄]) 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¡ÁC₄])=C₂⁹¨’C₄³. By Theorem 2.1,
K₁($\mathbb{Z}$[C₄¡ÁC₄],2$\mathbb{Z}$[C₄¡ÁC₄])=C₂³. By Theorem 5.5 in Ref.^[1],we have K₁($\mathbb{Z}$[C₄¡ÁC₄])=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¡ÁC₄]) is at least 1 and the 2-rank is at least 10. ¡õ
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