Why do we not have any subgroup of ${\mathbb Z}/{4 \mathbb Z}$ which maps isomorphically to the quotient group

abstract-algebragroup-theoryquotient-groupquotient-spacesvector-spaces

I have recently started watching the video lectures of the Harvard University Extension School regarding abstract algebra taught by Benedict Gross. I am now in lecture 9 where various results of vector spaces are being taught. At the end of the lecture the instructor has introduced the concept of quotient vector space. He showed that for a subspace $W$ of a vector space $V$ there is a subspace of the vector space say $W'$ such that $W'$ maps isomorphically to $V/W$. But he cautioned with a Bourbaki sign that this result is not necessarily valid for groups. For instance he took $G={\mathbb Z} / {4 \mathbb Z}$ and $H = {2 \mathbb Z} /{4 \mathbb Z}$. Then clearly $${\mathbb Z } /{2 \mathbb Z} \simeq {2 \mathbb Z}/{4 \mathbb Z} \subset {\mathbb Z}/{4 \mathbb Z}.$$ But he claimed that there does not exist any subgroup $H'$ of $G$ which maps isomorphically to $G/H$. I don't understand this part of this lecture. Why is it so happening?

Please help me in this regard.

Thank you very much.

Best Answer

Well, G/H is $\mathbb Z/2\mathbb Z$ so if H′ maps isomorphically to the quotient, then there needs to be an element of order 2 not in the kernel. But there is only one element of order 2 and it is in the kernel.

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