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Let $2m=2^{r_1}+2^{r_2}+...+2^{r_t}$. Then $T=W_{r_1}\times W_{r_2}\times ...\times W_{r_k}$ is a Sylow 2-subgroup of $O^\epsilon(2m,q)$.

Denote by $Z$ the center of $O^\epsilon(2m,q)$. Define $\phi_i=\Phi_i|_T$. Let $S'={\ker \phi_1}\cap {\rm ker \phi_2}$. Then $S'$ is a Sylow $2$-subgroups of $\Omega^{\epsilon}(2m,q)$. Since th determinant and spinorial norm of members in $Z$ are $1$ and perfect squares respectively, $Z\leq S'={\rm ker \phi_1}\cap {\rm ker \phi_2}$.

$T=S'W_{r_i}$ for all $i$.

Let $2m=2^{r_1}+2^{r_2}+...+2^{r_t}$. $T=W_{r_1}\times W_{r_2}\times ...\times W_{r_k}$ is a Sylow 2-subgroup of $O^\epsilon(2m,q)$.

Denote by $Z$ the center of $O^\epsilon(2m,q)$. Define $\phi_i=\Phi_i|_T$. Let $S'={\ker \phi_1}\cap {\rm ker \phi_2}$. Then $S'$ is a Sylow $2$-subgroups of $\Omega^{\epsilon}(2m,q)$. Since th determinant and spinorial norm of members in $Z$ are $1$ and perfect squares respectively, $Z\leq S'={\rm ker \phi_1}\cap {\rm ker \phi_2}$.

Let $2m=2^{r_1}+2^{r_2}+...+2^{r_t}$. Then $T=W_{r_1}\times W_{r_2}\times ...\times W_{r_k}$ is a Sylow 2-subgroup of $O^\epsilon(2m,q)$.

Denote by $Z$ the center of $O^\epsilon(2m,q)$. Define $\phi_i=\Phi_i|_T$. Let $S'={\ker \phi_1}\cap {\rm ker \phi_2}$. Then $S'$ is a Sylow $2$-subgroups of $\Omega^{\epsilon}(2m,q)$. Since th determinant and spinorial norm of members in $Z$ are $1$ and perfect squares respectively, $Z\leq S'={\rm ker \phi_1}\cap {\rm ker \phi_2}$.

$T=S'W_{r_i}$ for all $i$.

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Yi Wang
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Let $2m=2^{r_1}+2^{r_2}+...+2^{r_t}$. $T=W_{r_1}\times W_{r_2}\times ...\times W_{r_k}$ is a Sylow 2-subgroup of $O^\epsilon(2m,q)$.

Denote by $Z$ the center of $O^\epsilon(2m,q)$. Define $\phi_i=\Phi_i|_T$. Let $S'={\ker \phi_1}\cap {\rm ker \phi_2}$. Then $S'$ is a Sylow $2$-subgroups of $\Omega^{\epsilon}(2m,q)$. Since th determinant and spinorial norm of members in $Z$ are $1$ and perfect squares respectively, $Z\leq S'={\rm ker \phi_1}\cap {\rm ker \phi_2}$.

Denote by $Z$ the center of $O^\epsilon(2m,q)$. Define $\phi_i=\Phi_i|_T$. Let $S'={\ker \phi_1}\cap {\rm ker \phi_2}$. Then $S'$ is a Sylow $2$-subgroups of $\Omega^{\epsilon}(2m,q)$. Since th determinant and spinorial norm of members in $Z$ are $1$ and perfect squares respectively, $Z\leq S'={\rm ker \phi_1}\cap {\rm ker \phi_2}$.

Let $2m=2^{r_1}+2^{r_2}+...+2^{r_t}$. $T=W_{r_1}\times W_{r_2}\times ...\times W_{r_k}$ is a Sylow 2-subgroup of $O^\epsilon(2m,q)$.

Denote by $Z$ the center of $O^\epsilon(2m,q)$. Define $\phi_i=\Phi_i|_T$. Let $S'={\ker \phi_1}\cap {\rm ker \phi_2}$. Then $S'$ is a Sylow $2$-subgroups of $\Omega^{\epsilon}(2m,q)$. Since th determinant and spinorial norm of members in $Z$ are $1$ and perfect squares respectively, $Z\leq S'={\rm ker \phi_1}\cap {\rm ker \phi_2}$.

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Yi Wang
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Let $S$ be a Sylow $2$-subgroup of $P\Omega^\epsilon(2m,q)$, where $m\geq 4$, $q$ is a power of an odd prime and $q^m\equiv \epsilon (\rm mod~4)$. 

Let $F_q$ be the field of $q$ elements. Let $\Phi_1$ be the determinant mapping and $\Phi_2$ be the spinorial norm mapping $\Phi: O^\epsilon\rightarrow F_q^\times/{F_q^\times}^2\cong C_2$$\Phi_2: O^\epsilon\rightarrow F_q^\times/{F_q^\times}^2\cong C_2$. It is clear that

Let $S$ be a Sylow $2$-subgroup of $P\Omega^\epsilon(2m,q)$, where $m\geq 4$, $q$ is a power of an odd prime and $q^m\equiv \epsilon (\rm mod~4)$. Let $F_q$ be the field of $q$ elements. Let $\Phi_1$ be the determinant mapping and $\Phi_2$ be the spinorial norm mapping $\Phi: O^\epsilon\rightarrow F_q^\times/{F_q^\times}^2\cong C_2$. It is clear that

Let $S$ be a Sylow $2$-subgroup of $P\Omega^\epsilon(2m,q)$, where $m\geq 4$, $q$ is a power of an odd prime and $q^m\equiv \epsilon (\rm mod~4)$. 

Let $F_q$ be the field of $q$ elements. Let $\Phi_1$ be the determinant mapping and $\Phi_2$ be the spinorial norm mapping $\Phi_2: O^\epsilon\rightarrow F_q^\times/{F_q^\times}^2\cong C_2$. It is clear that

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