# Exercise 10.34 of Nielsen and Chuang : $-I$ not element of stabilizer group iff $g_j^2=I$ and $g_j \neq I$ where $g_j$ generators

I am stuck with this exercice of Nielsen and Chuang:

Let $$S = \langle g1,... ,gl \rangle$$.Show that $$−I$$ is not an element of S if and only if $$g^2_j = I$$ for all $$j$$,and $$g_j \neq − I$$ for all $$j$$.

$$S$$ is a subgroup of $$G_n$$ the set of $$n$$ Pauli matrices.

What I don't understand is that if I consider:

$$S=\langle X, -X \rangle$$

I have the two properties: $$g^2_j = I$$ for all $$j$$,and $$g_j \neq − I$$ for all $$j$$

However, $$X*(-X)=-X^2=-I$$

Thus it is a counter example of the property I should show. So I guess there is something I miss here but I don't get what...

• Is it possible that they are requiring that the generators be linearly independent, so that (for instance) their representation by bit-vectors (representing powers of $X$ and $Z$) are linearly independent? – Niel de Beaudrap Aug 28 '19 at 12:23
• @NieldeBeaudrap I would be surprised if so because he really introduces the notion of linearly independent generators just after this exercice. Even though he briefly talked about what it means. Furthermore I am not sure to understand because $X$ and $-X$ are linearly independent generator as $\langle X \rangle \neq \langle X, -X \rangle$ ? – StarBucK Aug 28 '19 at 12:29

Consulting my local copy of Nielsen & Chuang (10th anniversary edition, p. 457), the complete statement of the exercise is pretty much exactly as you have given it:

Exercise 10.34. Let $$\langle g_1, \ldots, g_l \rangle$$. Show that $$-I$$ is not an element of $$S$$ if and only if $$g_j^2 = I$$ for all $$j$$, and $$g_j \ne - I$$ for all $$j$$.

Reading the text as though I were learning the material for the first time, the context does not seem to impose many further restrictions. It is clear from context that each $$g_j$$ is meant to be a Pauli operator (an element of $$G_n$$ in the notation of Chapter 10) — and, just plausibly (but not very explicitly), we could suppose that the operators $$g_j$$ are meant to commute with one another. But:

• Without the explicit constraint that the operators commute, we could reasonably consider $$\langle X, Z \rangle$$ to be another counterexample, which contains the element $$-I = ZXZX$$.

• Even supposing that we restrict ourselves to abelian subgroups of $$G_n$$, the preceding text does not require that the generators $$g_j$$ be linearly independent operators, i.e. in this case that $$g_j \ne - g_k$$ for all $$j,k$$. This allows you to obtain the counterexample $$\langle X, -X \rangle$$ as you have done.

• Worse, it is not enough to simply check pairs of generators to see whether they are proportional to one another: the group $$\langle X\!\otimes\!X, Y\!\otimes\!Y, Z\!\otimes\!Z \rangle$$ has independent generators and linearly independent generators, but still include $$-I = (X\!\otimes\!X)(Y\!\otimes\!Y)(Z\!\otimes\!Z)$$.

I would propose that the exercise should be re-worded as follows:

Exercise 10.34  ( revised ). Let $$\langle g_1, \ldots, g_l \rangle$$ be a commutative subgroup of $$G_n$$. Show that $$-I$$ is not an element of $$S$$ if and only if $$g_j^2 = I$$ for all $$j$$, and no product of any subset of $$\{g_1, \ldots, g_l\}$$ is equal to $$-I$$.

It might seem as though the final condition (on subsets of the generators) rather gives the game away. However, from the fact that we are requiring that $$g_j^2 = I$$, we are essentially considering the Pauli group as a non-abelian group which contains elements of order $$2$$ and order $$4$$: from the very elementary starting point taken by Nielsen and Chuang, it is perhaps a useful exercise to take the time to show that in an abelian group whose generators all have order $$2$$, then every element $$g$$ is characterised by a subset of generators which you use exactly once in a product to obtain $$g$$. To people familiar with group theory (or whose prejudices are sufficiently steeped in computer science that they quickly leap to thinking about bit-vectors anyway), it is clear that the second condition amounts to saying the $$-I$$ is not generated by any product of generators — but we should recall that the target audience also includes physicists who might not see these things as quickly, as they have likely been pre-occupied mostly with solving differential equations or coding up numerical simulations before encountering Nielsen and Chuang.

This revised version of Exercise 10.34 is thus a relatively elementary exercise which gets you to think of subgroups of the Pauli group in which every element has order 2, getting you ready to think about them in terms of bit-vectors. It could be that this is actually be what Nielsen and Chuang originally intended. Whether or not this is the case, however, it is also the smallest modification of the exercise that I can think of which makes it correct.