Using tab delimiter in Cut in Unix Shell Scripting

cut -f 1 input.txt

This gives you the first column from the tab-delimited file input.txt. The default field delimiter for cut is the tab character, so there's no need to further specify this.

If the delimiter is actually a space, use

cut -d ' ' -f 1 input.txt

If it turns out that there are multiple tabs and/or spaces, use awk:

awk '{ print $1 }' input.txt

The shell loop is not necessary for this operation, regardless of whether you use cut or awk.

See also "Why is using a shell loop to process text considered bad practice?".

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The reason your script does not work is because the tab disappears when you echo the unquoted variable.

Related:

• Why is printf better than echo?

• Security implications of forgetting to quote a variable in bash/POSIX shells

Tab is the default separator for cut, you don't need an explicit argument for it.

However, you need to quote your variable to prevent the tabs from being turned into space.

SN=`echo "${line}"|cut -f1`

But you can also avoid using cut in the first place. Just set IFS to \t.

IFS=$'\t'
while read -r SN rest
do 
    echo "$SN"
done < input.txt