feat: fixed links

.gitignore

@@ -17,10 +17,11 @@ package-lock.json
 
 # IDE configurations
 .idea
-.vscode
+.vscode/*
 !.vscode/settings.json
 !.vscode/extensions.json
+!.vscode/tasks.json
 
 # Misc
-_sass/dist
+_sass/vendors
 assets/js/dist

Gemfile

@@ -2,7 +2,7 @@
 
 source "https://rubygems.org"
 
-gem "jekyll-theme-chirpy", "~> 7.2", ">= 7.2.4"
+gem "jekyll-theme-chirpy", "~> 7.3", ">= 7.3.1"
 
 gem "html-proofer", "~> 5.0", group: :test
 

_layouts/post.html

@@ -10,7 +10,9 @@ script_includes:
   - comment
 ---
 
-{% include lang.html %} {% include toc-status.html %}
+{% include lang.html %}
+
+{% include toc-status.html %}
 
 <article class="px-1" data-toc="{{ enable_toc }}">
   <header>
@@ -22,36 +24,40 @@ script_includes:
     <div class="post-meta text-muted">
       <!-- published date -->
       <span>
-        {{ site.data.locales[lang].post.posted }} {% include datetime.html
+        {{ site.data.locales[lang].post.posted }}
-        date=page.date tooltip=true lang=lang %}
+        {% include datetime.html date=page.date tooltip=true lang=lang %}
       </span>
 
       <!-- lastmod date -->
       {% if page.last_modified_at and page.last_modified_at != page.date %}
         <span>
-        {{ site.data.locales[lang].post.updated }} {% include datetime.html
+          {{ site.data.locales[lang].post.updated }}
-        date=page.last_modified_at tooltip=true lang=lang %}
+          {% include datetime.html date=page.last_modified_at tooltip=true lang=lang %}
         </span>
       {% endif %}
 
       <div class="d-flex justify-content-between">
         <!-- author(s) -->
         <span>
-          {% if page.author %} {% assign authors = page.author %} {% elsif
+          {% if page.author %}
-          page.authors %} {% assign authors = page.authors %} {% endif %} {{
+            {% assign authors = page.author %}
-          site.data.locales[lang].post.written_by }}
+          {% elsif page.authors %}
+            {% assign authors = page.authors %}
+          {% endif %}
+
+          {{ site.data.locales[lang].post.written_by }}
 
           <em>
-            {% if authors %} {% for author in authors %} {% if
+            {% if authors %}
-            site.data.authors[author].url -%}
+              {% for author in authors %}
-            <a href="{{ site.data.authors[author].url }}"
+                {% if site.data.authors[author].url -%}
-              >{{ site.data.authors[author].name }}</a
+                  <a href="{{ site.data.authors[author].url }}">{{ site.data.authors[author].name }}</a>
-            >
+                {%- else -%}
-            {%- else -%} {{ site.data.authors[author].name }} {%- endif %} {%
+                  {{ site.data.authors[author].name }}
-            unless forloop.last %}{{ '</em
+                {%- endif %}
-          >,
+                {% unless forloop.last %}{{ '</em>, <em>' }}{% endunless %}
-          <em
+              {% endfor %}
-            >' }}{% endunless %} {% endfor %} {% else %}
+            {% else %}
               <a href="{{ site.social.links[0] }}">{{ site.social.name }}</a>
             {% endif %}
           </em>
@@ -59,8 +65,7 @@ script_includes:
 
         <div>
           <!-- pageviews -->
-          {% if site.pageviews.provider and
+          {% if site.pageviews.provider and site.analytics[site.pageviews.provider].id %}
-          site.analytics[site.pageviews.provider].id %}
             <span>
               <em id="pageviews">
                 <i class="fas fa-spinner fa-spin small"></i>
@@ -77,37 +82,22 @@ script_includes:
   </header>
 
   {% if enable_toc %}
-  <div
+    <div id="toc-bar" class="d-flex align-items-center justify-content-between invisible">
-    id="toc-bar"
-    class="d-flex align-items-center justify-content-between invisible"
-  >
       <span class="label text-truncate">{{ page.title }}</span>
       <button type="button" class="toc-trigger btn me-1">
         <i class="fa-solid fa-list-ul fa-fw"></i>
       </button>
     </div>
 
-  <button
+    <button id="toc-solo-trigger" type="button" class="toc-trigger btn btn-outline-secondary btn-sm">
-    id="toc-solo-trigger"
+      <span class="label ps-2 pe-1">{{- site.data.locales[lang].panel.toc -}}</span>
-    type="button"
-    class="toc-trigger btn btn-outline-secondary btn-sm"
-  >
-    <span class="label ps-2 pe-1"
-      >{{- site.data.locales[lang].panel.toc -}}</span
-    >
       <i class="fa-solid fa-angle-right fa-fw"></i>
     </button>
 
     <dialog id="toc-popup" class="p-0">
-    <div
+      <div class="header d-flex flex-row align-items-center justify-content-between">
-      class="header d-flex flex-row align-items-center justify-content-between"
-    >
         <div class="label text-truncate py-2 ms-4">{{- page.title -}}</div>
-      <button
+        <button id="toc-popup-close" type="button" class="btn mx-1 my-1 opacity-75">
-        id="toc-popup-close"
-        type="button"
-        class="btn mx-1 my-1 opacity-75"
-      >
           <i class="fas fa-close"></i>
         </button>
       </div>
@@ -115,7 +105,9 @@ script_includes:
     </dialog>
   {% endif %}
 
-  <div class="content">{{ content }}</div>
+  <div class="content">
+    {{ content }}
+  </div>
 
   <div class="post-tail-wrapper text-muted">
     <!-- categories -->
@@ -123,11 +115,9 @@ script_includes:
       <div class="post-meta mb-3">
         <i class="far fa-folder-open fa-fw me-1"></i>
         {% for category in page.categories %}
-      <a
+          <a href="{{ site.baseurl }}/categories/{{ category | slugify | url_encode }}/">{{ category }}</a>
-        href="{{ site.baseurl }}/categories/{{ category | slugify | url_encode }}/"
+          {%- unless forloop.last -%},{%- endunless -%}
-        >{{ category }}</a
+        {% endfor %}
-      >
-      {%- unless forloop.last -%},{%- endunless -%} {% endfor %}
       </div>
     {% endif %}
 
@@ -147,16 +137,21 @@ script_includes:
     {% endif %}
 
     <div
-      class="post-tail-bottom d-flex justify-content-between align-items-center mt-5 pb-2"
+      class="
+        post-tail-bottom
+        d-flex justify-content-between align-items-center mt-5 pb-2
+      "
     >
       <div class="license-wrapper">
-        {% if site.data.locales[lang].copyright.license.template %} {% capture
+        {% if site.data.locales[lang].copyright.license.template %}
-        _replacement %}
+          {% capture _replacement %}
         <a href="{{ site.data.locales[lang].copyright.license.link }}">
           {{ site.data.locales[lang].copyright.license.name }}
         </a>
-        {% endcapture %} {{ site.data.locales[lang].copyright.license.template |
+        {% endcapture %}
-        replace: ':LICENSE_NAME', _replacement }} {% endif %}
+
+          {{ site.data.locales[lang].copyright.license.template | replace: ':LICENSE_NAME', _replacement }}
+        {% endif %}
       </div>
 
       {% include post-sharing.html lang=lang %}

_posts/mathematics/2025-08-22-group-structure-of-z-2n-z-star.md

@@ -0,0 +1,146 @@
+---
+share: true
+toc: true
+math: true
+categories:
+  - Mathematics
+path: _posts/mathematics
+tags:
+  - math
+  - cryptography
+title: Group Structure of $(\mathbb{Z}/2^n \mathbb{Z})^*$
+date: 2025-08-22
+github_title: 2025-08-22-group-structure-of-z-2n-z-star
+---
+
+To compute the rotation automorphism homomorphically, we use the fact that $(\Z/2^n\Z)^* \simeq \span{-1, 5}$. I couldn't find a clear proof of this result online, so I just accepted the fact although it wasn't very satisfying.
+
+After more than a year, I got a chance to revisit the rotation automorphism and I figured that I should clear things up once and for all. So I decided to compile a proof, drawn from many sources.
+
+## Theorem
+
+> **Theorem 1.** $(\Z/2^n \Z)^*$ is the direct product of a cyclic group of order $2$ and cyclic group of order $2^{n-2}$, for all $n \geq 2$.
+
+The above theorem is from Corollary 20 (2) of Section 9.5 in Abstract Algebra, 3rd Edition, Dummit and Foote.
+
+> **Theorem 2.** $(\Z/2^n\Z)^* \simeq \span{-1, 5}$ for $n \geq 3$.
+
+## Observations
+
+### Order of $5$ Modulo $2^n$
+
+> **Proposition.** $5^{2^{n-3}} \equiv 1 + 2^{n-1} \pmod {2^n}$ for $n \geq 3$.
+
+*Proof*. This is an easy proof with induction. Omitted.
+
+> **Lemma.** $5$ has order $2^{n-2}$ in $(\Z/2^n \Z)^*$, for $n \geq 2$.
+
+*Proof*. We will use strong induction. For $n = 2, 3$, the lemma can be checked by direct computation. Now assume that the order of $5$ is $2^{k-2}$ in $(\Z/2^k\Z)^*$, for all $3 \leq k \leq n$.
+
+Let $r$ be the order of $5$ modulo $2^{n+1}$. Then $2^{n-2} \mid r$. This is from the fact that
+
+$$
+5^r \equiv 1 \pmod {2^{n+1}} \implies 5^r \equiv 1 \pmod {2^n}.
+$$
+
+Therefore $r$ must be a multiple of $2^{n-2}$. (order of $5$ modulo $2^n$) But from the above proposition, $5^{2^{n-2}} \equiv 1 + 2^n \pmod {2^{n+1}}$, so $r \neq 2^{n-2}$. The next candidate of $r$ is $2^{n-1}$ since it should be a multiple of $2^{n-2}$. Observe that
+
+$$
+5^{2^{n-1}} = \paren{5^{2^{n-2}}}^2 = (1 + 2^{n})^2 \equiv 1 \pmod {2^{n+1}},
+$$
+
+completing the proof.
+
+### Group is Not Cyclic
+
+> **Proposition.** Let $G = \span{x}$ be a cyclic group of finite order $n < \infty$. For each divisor $a$ of $n$, there exists a unique subgroup of $G$ with order $a$.
+
+*Proof*. Since $a \mid n$, set $d = n /a$. Then $\span{x^d}$ is a subgroup of order $a$, showing existence.
+
+For uniqueness, suppose $H \neq \span{x^d}$ is another subgroup of $G$ with order $a$. Since subgroups of cyclic groups are also cyclic, $H = \span{x^k}$ where $k$ is the smallest positive integer with $x^k \in H$. Then from
+
+$$
+\frac{n}{d} = a = \abs{H} = \frac{n}{\gcd(n, k)},
+$$
+
+$d = \gcd(n, k)$. So $k$ is a multiple of $d$, resulting in $x^k \in \span{x^d}$. Therefore, $H \leq \span{x^d}$, but the two groups have the same order, so $H = \span{x^d}$.
+
+> **Lemma.** $(\Z/2^n \Z)^*$ is not cyclic for any $n \geq 3$.
+
+*Proof*. $(\Z/2^n\Z)^*$ has two distinct subgroups of order $2$. For $n \geq 3$,
+
+$$
+(2^n - 1)^2 \equiv (-1)^2 \equiv 1 \pmod {2^n}
+$$
+
+and
+
+$$
+(2^{n-1}-1)^2 = 2^{2n-2} - 2^n + 1 \equiv 1 \pmod {2^n}.
+$$
+
+Both $2^n-1$ and $2^{n-1} - 1$ have order $2$ modulo $2^n$ and they are distinct since $n \geq 3$. By the above proposition, $(\Z/2^n\Z)^*$ cannot be cyclic.
+
+## Proof of Theorem 1
+
+*Proof*. $(\Z/2^n\Z)^*$ is a finitely generated abelian group, so the fundamental theorem of finitely generated abelian groups applies here. We know that group has order $2^{n-1}$, and from the above results,
+
+- $(\Z/2^n \Z)^*$ has an element of order $2^{n-2}$,
+- $(\Z/2^n \Z)^*$ is not cyclic for $n \geq 3$.
+
+Thus, for $n \geq 3$, the only possible case is $(\Z/2^n\Z)^* \simeq \Z _ 2 \times \Z_{2^{n-2}}$. As for $n = 2$, $(\Z/4\Z)^* \simeq \Z_2 \times \Z_1$ is pretty obvious.
+
+*Note*. I'm still looking for an elementary proof that doesn't use the fundamental theorem. This sort of feels like nuking a mosquito.
+
+## More Observations
+
+> **Lemma.** Suppose that $H$ and $K$ are normal subgroups of $G$ and $H \cap K = \braces{1}$. Then $HK \simeq H \times K$.
+
+*Proof*. Construct an isomorphism $\varphi : H \times K \ra HK$ such that $(h, k) \mapsto hk$.
+
+Since $H, K \unlhd G$, observe that $hkh\inv k \inv \in K \cap H = \braces{1}$ and $hk = kh$. Therefore,
+
+$$
+\varphi(h, k) \cdot \varphi(h',k') = hkh'k' = hh' kk' = \varphi\paren{(h, k)\cdot (h', k')}
+$$
+
+and $\varphi$ is a homomorphism.
+
+Next, if $\varphi(h, k) = hk = 1$, we have $h = k\inv \in H\cap K = \braces{1}$. Then $h = k = 1$, showing that $\ker \varphi$ is trivial and $\varphi$ is injective.
+
+Surjectivity of $\varphi$ is trivial. $\varphi$ is an isomorphism and $HK \simeq H \times K$.
+
+> **Proposition.** As subgroups of $(\Z/2^n\Z)^*$, $\span{-1} \cap \span{5} = \braces{1}$ for $n \geq 3$.
+
+*Proof*. It suffices to show that $-1 \notin \span{{5}}$. Suppose that $-1 \in \span{5}$. Since $\span{5}$ is cyclic, it has a unique element of order $2$. Since $5$ has order $2^{n-2}$, it must be the case that $-1 \equiv 5^{2^{n-3}} \pmod {2^n}$.
+
+Then we have
+
+$$
+-1 \equiv 5^{2^{n-3}} \equiv 1 + 2^{n-1} \pmod {2^n},
+$$
+
+which gives $2^{n-1} + 2 \equiv 0 \pmod {2^n}$. But for $n \geq 3$, this is impossible since $0 < 2^{n-1} + 2 < 2^n$. Contradiction.
+
+*Note*. If $-1 \in \span{5}$, then maybe $5$ would have generated the whole group. But the group isn't cyclic, so we have a contradiction?
+
+## Proof of Theorem 2
+
+*Proof*. Since we are dealing with commutative groups, all subgroups are normal. We have $\span{-1}, \span{5} \unlhd (\Z/2^n\Z)^*$ and $\span{-1} \cap \span{5} = \braces{1}$. Therefore,
+
+$$
+(\Z/2^n\Z)^* \simeq \Z_2 \times \Z_{2^{n-2}} = \span{-1} \times \span{5} \simeq \span{-1}\span{5}.
+$$
+
+This means that we can uniquely write all elements of $(\Z/2^n\Z)^*$ as $(-1)^a 5^b$ for ${} 0 \leq a < 2 {}$, $0 \leq b < 2^{n-2}$. From commutativity, this exactly equals the subgroup generated by $-1$ and $5$, which is $\span{-1, 5}$. This concludes the proof.
+
+## Notes
+
+The theorem wasn't so trivial after all, but I'm still happy to have resolved a long overdue task.
+
+## References
+
+- My notes taken from abstract algebra class
+- <https://math.stackexchange.com/q/459815>
+- <https://math.stackexchange.com/q/3881641>
+- <https://math.stackexchange.com/a/4910312/329909>

assets/css/jekyll-theme-chirpy.scss

@@ -1,7 +1,7 @@
 ---
 ---
 
-@import 'main
+@use 'main
 {%- if jekyll.environment == 'production' -%}
   .bundle
 {%- endif -%}