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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
Thanks a lot! Is what you mean when $Z({\rm Spin}_n^\epsilon(q))\cong C_4$, then, for question A, all geneartors of order 4 are not squares, and all generators of order 2 are yes, and for question $B$, the answer is also negative; when $Z({\rm Spin}_n^\epsilon(q))\cong C_2\times C_2$, then for question A, all generators of order 2 are squares, and for question B, the answer is also yes?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
Thanks for your comments, sorry for my behavior, I will not do any further edits.
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?
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