In working on a paper, I found myself in the amusing but unusual situation of having a group $latex G$, a subgroup $latex H$ and an element $latex \\xi\\in G$ such that
\n$latex \\xi H\\xi =H.$<\/p>\n
This certainly can happen: the two obvious cases are when $latex H=G$, or when $latex \\xi$ is an involution that happens to be in the normalizer $latex N(H)$ of $latex H$. <\/p>\n
In fact the general case is just a tweak of this last case: we have $latex \\xi H\\xi=H$ if and only if $latex \\xi^2\\in H$ and $latex \\xi\\in N(H)$, or in other words, if $latex \\xi$ belongs to the normalizer, and is an involution modulo $latex H$.<\/p>\n
This is of course easy to check. I then asked myself: what about the possibility that
\n$latex \\xi^a H\\xi^b = H$,
\nwhere $latex a$ and $latex b$ are arbitrary integers? Can one classify when this happens? The answer is another simple exercise (that I will probably use when I teach Algeba I next semester): this is the case if and only if $latex \\xi^{a+b}\\in H$ and $latex \\xi^{(a,b)}\\in N(H)$, where $latex (a,b)$ is the gcd of $latex a$ and $latex b$. In particular, for all pairs $latex (a,b)$ where $latex a$ and $latex b$ are coprime, the condition above implies that $latex \\xi$ belongs to the normalizer of $latex H$.<\/p>\n
Here is the brief argument: having fixed $latex \\xi$, let
\n$latex M=\\{(\\alpha,\\beta)\\in\\mathbf{Z}^2\\,\\mid\\, \\xi^{\\alpha}H\\xi^{\\beta} =H\\}.$
\nThis set is easily seen to be a subgroup of $latex \\mathbf{Z}^2$. Furthermore, note that $latex (\\alpha,\\beta)\\in M$ implies that $latex \\xi^{\\alpha+\\beta}\\in H$, which in turns means that $latex (\\alpha+\\beta,0)\\in M$ and $latex (0,\\alpha+\\beta)\\in M$.<\/p>\n
Hence if $latex (a,b)\\in M$, we get
\n$latex (a,-a)=(a,b)-(0,a+b)\\in M,\\quad\\quad (b,-b)=(a+b,0)-(a,b)\\in M$,
\nso that
\n$latex M\\cap (1,-1)\\mathbf{Z}$
\ncontains $latex (a,-a)\\mathbf{Z}\\cap (b,-b)\\mathbf{Z}$, which is just
\n$latex (d,-d)\\mathbf{Z},$
\nwhere $latex d=(a,b)$. But $latex (d,-d)\\in M$ means exactly that $latex \\xi^{d}\\in N(H)$.<\/p>\n
Thus we have got the first implication. Conversely, the conclusion means exactly that
\n$latex (a+b,0)\\in M,\\quad (d,-d)\\in M.$
\nBut then
\n$latex (a,b)=(a+b,0)-(b,-b)=(a+b,0)-b\/d (d,-d)\\in M$
\nshows that $latex \\xi^a H\\xi^b=H$.<\/p>\n
To finish, how did I get to this situation? This can arise quite naturally as follows: one has a collection $latex X$ of representations $latex \\rho$ of a fixed group $latex \\Gamma$, and an action of a group latex $latex G$ on these representations $latex X$ (action up to isomorphism really).
\nFor a given representation $latex \\rho_0$, we can then define a group
\n$latex H=\\{g\\in G\\,\\mid\\, g\\cdot \\rho_0\\simeq \\rho_0\\}$
\nand also a subset
\n$latex T=\\{g\\in G\\,\\mid\\, g\\cdot \\rho_0\\simeq D(\\rho_0)\\},$
\nwhere $latex D(\\cdot)$ denotes the contragredient representation.<\/p>\n
It can be that $latex T$ is empty, but let us assume it is not. Then $latex T$ has two properties: (1) it is a coset of $latex H$ (because $latex H$ acts on $latex T$ simply transitively); (2) we have $latex g^2\\in H$ for all $latex g\\in T$ (because the contragredient of the contragredient is the representation itself).<\/p>\n
This means that, for some $latex \\xi\\in G$, we have $latex T=\\xi H$, and furthermore
\n$latex \\xi H\\xi=H$
\nsince $latex \\xi g\\xi g\\in H$ for all $latex g\\in H$. <\/p>\n
By the previous discussion, we therefore get a good handle on the structure of $latex T$: either it is empty, or it is of the form $latex \\xi H$ for some $latex \\xi\\in G$ such that $latex \\xi^2\\in H$ and $latex \\xi$ normalizes $latex H$ in $latex G$. In particular, if $latex H$ is trivial (which happens often), either $latex T$ is empty, or it consists of a single element $latex \\xi$ which is an involution of $latex G$.<\/p>\n","protected":false},"excerpt":{"rendered":"
In working on a paper, I found myself in the amusing but unusual situation of having a group $latex G$, a subgroup $latex H$ and an element $latex \\xi\\in G$ such that $latex \\xi H\\xi =H.$ This certainly can happen: the two obvious cases are when $latex H=G$, or when $latex \\xi$ is an involution … Continue reading Normalizers everywhere<\/span><\/a><\/p>\n","protected":false},"author":625,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1086,2],"tags":[],"class_list":["post-4258","post","type-post","status-publish","format-standard","hentry","category-exercise","category-blogroll"],"_links":{"self":[{"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/posts\/4258","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/users\/625"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/comments?post=4258"}],"version-history":[{"count":0,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/posts\/4258\/revisions"}],"wp:attachment":[{"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/media?parent=4258"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/categories?post=4258"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/tags?post=4258"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}